quadratic equations - Testlabz.com

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QUADRATIC EQUATIONS

A. SUMMATIVE ASSESSMENT

4

4.1 QUADRATIC EQUATIONS

1. An equation of the form 2 0 ax bx c ,where a, b, c are real numbers and a ≠ 0, is calledthe standard form of a quadratic equation in x.

2. A real number is called a root of the

quadratic equation 2 0 ax bx c , 0a if2 0. a b c Any quadratic equation can

have at most two roots.

Note : If is a root of 2 0ax bx c ,

then we say that(i) x satisfies the equation

ax2 + bx + c = 0 or(ii) x is a solution of the equation2 0ax bx c 3. The roots of a quadratic equation

2 0+ + =ax bx c are called the zeroes of the

polynomial 2ax bx c .4. Solving a quadratic equation means

finding its roots.

TEXTBOOK’S EXERCISE 4.1

Q.1. Check whether the following are quadraticequations :

(i) 2( 1) 2( 3) x x

(ii) 2 2 ( 2)(3 ) x x x

(iii) ( 2)( 1) ( 1)( 3) x x x x

(iv) ( 3)(2 1) ( 5) x x x x

(v) (2 1)( 3) ( 5)( 1) x x x x

(vi) 2 23 1 ( 2) x x x

(vii) 3 2( 2) 2 ( 1) x x x

(viii) 3 2 34 1 ( 2) x x x x

Sol. Standard form of quadratic equation is2 0; ax bx c 0a .

(i) 2( 1) 2( 3)+ = −x x

2 1 2 2( 3)+ + = −x x x

2 1 2 2 6+ + = −x x x

2 1 2 2 6 0+ + − + =x x x 2 7 0+ =x

It is of the form2 0,+ + =ax bx c

where a = 1, b = 0 and c = 7.Hence, it is a quadratic equation.

(ii) 2 2 ( 2) (3 )x x x− = − − 2 2 6 2− = − +x x x

2 2 2 6 0x x x− − + = 2 4 6 0x x− + =

It is of the form 2 0+ + =ax bx c ,

where 1,a = 4b = − and 6=c .

Hence, it is a quadratic equation.

(iii) ( 2)( 1) ( 1)( 3)− + = − +x x x x

( 1) 2( 1) ( 3) 1( 3)+ − + = + − +x x x x x x

2 22 2 3 3+ − − = + − −x x x x x x

2 22 2 3− − = + −x x x x

2 22 3 2 0+ − − + + =x x x x

3 1 0− =x

It is not of the form 2 0 ax bx c .Hence, it is not a quadratic equation.

(iv) ( 3)(2 1) ( 5) x x x x

2(2 1) 3(2 1) 5+ − + = +x x x x x

2 22 6 3 5+ − − = +x x x x x

2 22 5 3 5 0− − − − =x x x x

2 10 3 0− − =x xIt is of the form 2 0+ + =ax bx c , where

1, 10= = −a b and 3= −c .


Question Bank In Mathematics Class X (Term–II)

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(v) (2 1) ( 3) ( 5) ( 1)− − = + −x x x x

2 ( 3) 1( 3) ( 1) 5( 1)− − − = − + −x x x x x x

2 22 6 3 5 5− − + = − + −x x x x x x

2 22 7 3 4 5 x x x x

2 22 7 3 4 5 0− + − − + =x x x x

2 11 8 0− + =x x

It is of the form 2 0+ + =ax bx c , where 1,a =11b = − and 8=c .


(vi) 2 23 1 ( 2)+ + = −x x x

2 23 1 4 4+ + = + −x x x x

2 23 1 4 4 0+ + − − + =x x x x 7 3 0− =x

It is not of the form 2 0+ + =ax bx c . Hence, thegiven equation is not a quadratic equation.

(vii)3 2( 2) 2 ( 1)+ = −x x x

3 3 3(2) 3 2 ( 2) 2 2 x x x x x

3 38 6 ( 2) 2 2+ + + = −x x x x x

3 2 38 6 12 2 2+ + + = −x x x x x

3 2 38 6 12 2 2 0+ + + − + =x x x x x

3 26 14 8 0− + + + =x x x

It is not of the form 2 0+ + =ax bx c . Hence, thegiven equation is not a quadratic equation.

(viii) 3 2 34 1 ( 2)− − + = −x x x x

3 2 3 34 1 (2)x x x x− − + = − 3 2 ( 2)x x− × × −

x3 2 34 1 8 6 ( 2) x x x x x

3 3 3[( ) 3 ( )] a b a b ab a b

3 2 3 24 1 8 6 12− − + = − − +x x x x x x

3 2 3 24 1 8 6 12 0 x x x x x x

22 13 9 0− + =x x

It is of the form 2 0+ + =ax bx c ,where a = 2, b = –13 and c = 9.

Hence, the given equation is a quadratic equation.

Q.2. Represent the following situations in theform of quadratic equations :

(i) The area of a rectangular plot is528 m2. The length of the plot (in metres) is onemore than twice its breadth. We need to find thelength and breadth of the plot.

(ii) The product of two consecutive positive

integers is 306. We need to find the integers.(iii) Rohan's mother is 26 years older than

him. The product of their ages (in years) 3 yearsfrom now will be 360. We would like to findRohan's present age.

(iv) A train travels a distance of 480 km at auniform speed. If the speed had been 8 km/h less,then it would have take 3 hours more to coverthe same distance. We need to find the speed ofthe train.

Sol. (i) Let breadth of rectangular plot = x m

Then, length of the rectangular plot= (2x + 1) mSo, area of rectangular plot

= [x (2x + 1)] m2 = (2x2 + x) m2

As per condition : 22 528+ =x x

22 528 0+ − =x xHence, given problem in the form of quadratic

equation is 22 528 0. x x

(ii) Let two consecutive positive integers are xand x + 1.

Product of integers = x (x + 1) = 2 +x x

As per condition : 2 306+ =x x

2 306 0+ − =x x

Hence, given problem in the form of quadraticequation is x2 + x – 306 = 0.

(iii) Let present age of Rohan = x yearsPresent age of Rohan's mother = (x + 26) yearsAfter 3 years, Rohan's age = (x + 3) yearsAfter 3 years, Rohan's mother's age= (x + 26 + 3) years = (x + 29) yearsSo, product = (x + 3) (x + 29)= x2 + 29x + 3x + 87= x2 + 32x + 87As per condition : x2 + 32x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x – 273 = 0

Hence, given problem in the form of quadraticequation is x2 + 32x – 273 = 0.

(iv) Let speed of train = x km/hourDistance covered by train = 480 km

Time taken by train =Distance 480

Speed=

xhour

If speed of train is decreased by 8 km/hour

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Then, new speed of train = (x – 8) km/hour

Time taken by train =480

8−xhour

As per condition :480

8−x–

480

x= 3

480 480( 8) 3

( 8) 1

− − =−

x x

x x

2

480 480 3840 3

18

− +=

−x x

x x 3840 = 3(x2 – 8x) 3x2 – 24x = 3840

x2 – 8x = 1280

x2 – 8x – 1280 = 0

Hence, given problem in the form of quadraticequation is x2 – 8x – 1280 = 0.

OTHER IMPORTANT QUESTIONS

Q.1. If p(x) = 0 is a quadratic equation, thenp(x) is a polynomial of degree :

(a) one (b) two (c) three (d) fourSol. (b) p(x) is a polynomial of degree 2.

Q.2. Which of the following is not aquadratic equation : [2011 (T-II)]

(a) (x – 2)2 + 1 = 2x – 3

(b) x(x + 1) + 8 = (x + 2)(x – 2)

(c) x(2x + 3) = x2 + 1

(d) (x + 2)3 = x3 – 4

Sol. (b) x(x + 1) + 8 = (x + 2)(x – 2)

x2 + x + 8 = x2 – 4 x + 12 = 0

It is not of the form ax2 + bx + c = 0. Hence, thegiven equation is not a quadratic equation.

Q.3. If 8 is a root of the equationx2 – 10x + k = 0, then the value of k is :

[2011 (T-II)](a) 2 (b) 8 (c) –8 (d) 16

Sol. (d) (8)2 – 10 × 8 + k = 0

64 – 80 = –k k = 16

Q.4. Which of the following is a root of theequation 2x2 – 5x – 3 = 0? [2011 (T-II)]

(a) x = 3 (b) x = 4

(c) x = 1 (d) x = –3

Sol. (a) 2(3)2 – 5(3) – 3 = 0

18 – 15 – 3 = 0

Q.5. Which one is not a quadratic equationin x?

(a) 22 7 14 0 x x

(b)25

3 06

x x

(c) 2 3 4 0 x x x

(d) 100x2 – 20x + 1 = 0

Sol. (c) 2 3 4 0x x x− − + = is not a quadratic

equation as it contains x .

Q.6. (x + 2)3 = x(x2 – 1) is a :

(a) biquadratic equation

(b) cubic equation

(c) quadratic equation

(d) linear equation

Sol. (c) (x + 2)3 = x(x2 – 1)

x3 + 12x + 6x2 + 8 = x3 – x

6x2 + 13x + 8 = 0

So, it is a quadratic equation.

Q.7. Which of the following is a root of theequation 3x2 – 2x – 1 = 0?

(a) x = –1 (b) x = –2

(c) x = 1 (d) x = 2

Sol. (c) 3(1)2 – 2 × 1 – 1 = 3 – 2 – 1 = 3 – 3 = 0

Q.8. x = 2 is a solution of the equation :

(a) 2 02 4 x x

(b) 2 2 4 0. x x

(c) 3x2 + 5x + 2 = 0

(d) (a) and (b) both

Sol. (a) 2( 2) 2 2 4 = 2 + 2 – 4 = 0

Q.9. For what value of k, x = 2 is a solutionof kx2 + 2x – 3 = 0?

(a) k =1

2 (b) k =

1

2

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(c) k =1

4 (d) k =

1

4Sol. (c) k(2)2 + 2 × 2 – 3 = 0

4k + 4 – 3 = 0 k =1

4−

Q.10. If x = 2 is a root of the equation3x2 – 2kx + 5 = 0, then k is equal to :

(a)4

17(b)

17

4(c)

1

17(d)

1

4Sol. (b) 3 × (2)2 – 2 × 2 × k + 5 = 0

12 – 4k + 5 = 0 k =17

4Q.11. Check whether the following equations

are quadratic or not :(i) (x – 1)(x – 3) = (x + 5)(x – 1)

(ii) x2 + 2x – 6 x + 5 = 0

(iii) x3 – 4x2 + 1 = (x – 2)3

Sol. (i) (x – 1) (x – 3) = (x + 5) (x – 1)

x2 – 4x + 3 = x2 + 4x – 5

–4x – 4x + 3 + 5 = 0

–8x + 8 = 0which is not of the form ax2 + bx + c = 0.Thus, (x – 1) (x – 3) = (x + 5) (x – 1) is not a quadratic

equation.

(ii) x2 + 2x – 6 x + 5 = 0

The given equation contains a term involving ,x

i.e., x1/2.Since, x2 + 2x – 6 5x + is not a quadratic

polynomial, therefore, x2 + 2x – 6 5x = 0 is not aquadratic equation.

(iii) x3 – 4x2 + 1 = (x – 2)3

x3 – 4x2 + 1 = x3 – 8 – 6x(x – 2) –4x2 + 1 = –8 – 6x2 + 12x –4x2 + 6x2 – 12x + 1 + 8 = 0 2x2 – 12x + 9 = 0,which is a quadratic equation as it is of the form

ax2 + bx + c = 0.

Thus, the given equation is a quadratic equation.

Q.12. Determine whether the given valuesare solutions (roots) of the equation or not.

(i) 2 2 4 0, 2, 2 2x x x x+ − = = = −

(ii) (3x + 8) (2x + 5) = 0, x =2

2 ,3

x =1

22

(iii) a2x2 – 3abx + 2b2 = 0, x = ,a b

xb a

=

(iv) 6x2 – x – 2 = 0, x =1 2

,2 3

x− =

Sol. (i) 2 2 4 0 x x

Putting x = 2 , we get

2( 2) ( 2)( 2) 4 0+ − = 2 + 2 – 4 = 0

4 – 4 = 0

0 = 0, which is true.

∴ x = 2 is a solution of the equation.

Now, putting x = 2 2− , we get

2( 2 2) 2( 2 2) 4 0− + − − = 8 – 4 – 4 = 0 0 = 0, which is true.

∴ x = 2 2 is a solution of the equation.

(ii) We have (3x + 8) (2x + 5) = 0

6x2 + 15x + 16x + 40 = 0 6x2 + 31x + 40 = 0

Putting x =2 8

23 3

= in the equation, we get

28 8

6 31 40 03 3

⋅ + ⋅ + =

384 248

40 09 3

+ + =

384 + 744 + 360 = 0

1488 = 0, which is not true.

Hence, x =2

23

is not a solution of the given

equation.

Putting x =1

22

=5

2in the given equation, we

get2

5 56 31 40 0

2 2

⋅ + ⋅ + =

150 155

404 2

+ + = 0

150 + 310 + 160 = 0

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620 = 0, which is not true.

Hence, x =1

22

is not a solution of the given

equation.

(iii) a2x2 – 3abx + 2b2 = 0

Putting x =a

bin the given equation, we get

22 23 2 0

a aa ab b

b b

− + =

4

2 22

3 2 0a

a bb

− + =

a4 – 3a2b2 + 2b4 = 0, which is not true.

∴ x =a

bis not a solution of the given equation.

Again, putting x =b

ain the given equation, we get

22 23 2 0

b ba ab b

a a

− + = b2 – 3b2 + 2b2 = 0 –2b2 + 2b2 = 0 0 = 0, which is true.

∴ x =b

ais a solution of the given equation.

(iv) 6x2 – x – 2 = 0

Putting x =1

2− in the given equation, we get

21 1

6 2 02 2

1 1

6 2 04 2

3 1 4

02 2

−+ =

3 3

0,2 2

− + = 0 = 0, which is true.

∴ x =1

2 is a solution of the given equation.

Again, putting x =2

,3

in the given equation, we

get

22 2

6 2 03 3

− − =

4 2

6 2 09 3

× − − = 8 2

2 03 3

− − =

8 2

2 03

−− =

2 – 2 = 0 0 = 0, which is true.

x =2

3is a solution of the given equation.

Q.13. In each of the following, find the valueof k for which the given value is a solution of thegiven equation : [Imp.]

(i) x2 – x(a + b) + k = 0, x = a

(ii) kx2 + 2x – 4 = 0, x = 2Sol. (i) Since x = a is a root of the given

equation. Therefore, it satisfies the equation.i.e., a2 – a (a + b) + k = 0 a2 – a2 – ab + k = 0 k = ab

(ii) Since x = 2 is a root of the given equation.

Therefore, it satisfies the equation.

i.e., 2( 2) 2 2 4 0k + ⋅ − =2k + 2 – 4 = 02k – 2 = 0 k = 1

Q.14. If x = – 2 and x =1

5− are solutions

of the equation 5x2 + kx + λ = 0, find the values

of k and λ . [Imp.]

Sol. Since x = –2 and x =1

5− are solutions of the

equation, 5x2 + kx + = 0

5(–2)2 + k(–2) + = 0

2k – = 20 …(i)

and2

1 15 0

5 5k

− + − + λ =

1

05 5

k− + λ =

1 5 0k− + λ = 5 1k − λ = …(ii)

From (i), we get 2 20kλ = −Substituting this value of λ in (ii), we get

5(2 20) 1k k− − = 10 100 1k k− + = –9k = –99 k = 11

= 2k – 20 = 2(11) – 20 = 2

Hence, k = 11 and = 2

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Q.15. If x =2

3and x = –3 are the roots of

the equation ax2 + 7x + b = 0, find the values ofa and b. [Imp.]

Sol. Since x =2

3is a root of ax2 + 7x + b = 0,

we have,

22 2

7 03 3

a b + + =

4 14

09 3

ab+ + = 4a + 42 + 9b = 0

4a + 9b + 42 = 0 ...(i)

Again, x = –3 being a root of ax2 + 7x + b = 0, we

have, a(–3)2 + 7(–3) + b = 0

9a – 21 + b = 0

9a + b – 21 = 0 ...(ii)

Solving (i) and (ii), we get

a = 3 and b = – 6

PRACTICE EXERCISE 4.1A

Choose the correct option (Q. 1 – 4) :1. Which one of the following is not a

quadratic equation?(a) (x – 2)2 = 2(x + 3)

(b) x2 + 3x = –1 (1 – 3x)2

(c) (x + 2)(x – 1) = x2 – 2x – 3

(d) x3 – x2 + 2x + 1 = (x + 1)3

2. Which of the following is a quadraticequation?

(a) x2 + 2x + 1 = (4 – x)2 + 3

(b) –2x2 = (5 – x)5

2

xx

(c) (k + 1)x2 +3

2x = 7, k = –1

(d) x3 – x2 = (x – 1)3

3. Which of the following equations has 2 asa root?

(a) x2 – 4x + 5 = 0 (b) x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0 (d) 3x2 – 6x – 2 = 0

4. If1

2is a root of the equation

x2 + kx –5

4= 0, then the value of k is :

(a) 2 (b) –2 (c)1

4(d)

1

25. Which of the following are quadratic

equations in x?

(i)1

1 xx

(ii)2

2

15 x

x

(iii)23

x xx

(iv) 22 3 9 0 x x

(v) 2 2 5 0 x x x(vi) 3x2 – 5x + 9 = x2 – 7x + 3

(vii) 2 22 7 8 5 7 x x x

(viii)

21 1

3 4

x xx x

(ix) (x – 2)2 + 1 = x2 – 3(x) x(2x + 3) = x2 + 1

(xi)

21

xx

= 6

(xii) x(x + 1) + 8 = (x + 2) (x – 2)

6. Determine whether x = 3 and x = 2 3

are solutions of the equation 2 3 3 6 0x x− + = .

7. Determine whether x =3

2and x =

4

3

−are

the solutions of the equation 6x2 – x – 12 = 0 or not.8. For the quadratic equation

2x2 – 5x – 3 = 0, determine which of thefollowing are its solutions?

(i) x = 3 (ii) x = –2

(iii) x =1

2 (iv) x =

1

3

9. For what values of a and b, x =3

4and

x = –2 are solutions of the equationax2 + bx – 6 = 0 ?

10. In each of the following, find the value ofk for which the given value is a solution of thegiven equation :

(i) 7x2 + kx – 3 = 0, x =2

3(ii) x2 + 3ax + k = 0, x = –a

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Q.1. Find the roots of the followingquadratic equations by factorisation :

(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0

(iii) 22 7 5 2 0 x x

(iv) 22 x x1

08=

(v) 100x2 – 20x + 1 = 0Sol. (i) x2 – 3x – 10 = 0 x2 – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5) (x + 2) = 0Therefore, x – 5 = 0 or x + 2 = 0i.e., x = 5 or x = –2Hence, 5 and –2 are the roots of the given

quadratic equation.

(ii) 2x2 + x – 6 = 0

2x2 + 4x – 3x – 6 = 0

2x(x + 2) – 3(x + 2) = 0

(x + 2) (2x – 3) = 0

x + 2 = 0 or 2x – 3 = 0,

x = –2 or x =3

2

Hence, –2 and3

2are the roots of the given

quadratic equation.

(iii) 22 7 5 2 0x x+ + = 22 2 5 5 2 0x x x+ + + =

2 ( 2) 5( 2) 0x x x+ + + =

( 2)( 2 5) 0x x+ + =

2 0x + = or 2 5 0 x

x = 2− or x =5

2−

Hence, 2− and5

2

−are the roots of the given

quadratic equation.

(iv) 2 12 0

8x x− + =

16x2 – 8x + 1 = 0 16x2 – 4x – 4x + 1 = 0 4x(4x – 1) – 1(4x – 1) = 0 (4x – 1) (4x – 1) = 0

x =1

4or x =

1

4

Hence,1

4and

1

4are the roots of the given

quadratic equation.

(v) 100x2 – 20x + 1 = 0 100x2 – 10x – 10x + 1 = 0 10x(10x – 1) – 1(10x – 1) = 0 (10x – 1) (10x – 1) = 0

x =1

10or x =

1

10

Hence, 1

10and

1

10are the roots of given

quadratic equation.

4.2 SOLUTION OF A QUADRATICEQUATION BY FACTORISATION

1. If ax2 + bx + c can be factorised as

( )( ),x x− α − β then ax2 + bx + c = 0 is

equivalent to ( )( ) 0x x− α − β = .

Thus, ( )( ) 0x x− α − β = 0x − α = or 0x − β =

i.e., x = ,α or x = .β

Here α and β are called the roots of the

equation ax2 + bx + c = 0.

2.To solve a quadratic equation by factorisa-tion :

(a) Clear fractions and brackets, ifnecessary.

(b) Transfer all the terms to L.H.S. andcombine like terms.

(c) Write the equation in the standard form,i.e.,

ax2 + bx + c = 0(d) Factorise the L.H.S.(e) Put each factor equal to zero and solve.(f) Check each value by substituting it in

the given equation.

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Q.2. Solve the problems :

(i) John and Jivanti together have 45marbles. Both of them lost 5 marbles each, andthe product of the number of marbles they nowhave is 124. We would like to find out how manymarbles they had to start with.

(ii) A cottage industry produces a certainnumber of toys in a day. The cost of productionof each toy (in rupees) was found to be 55 minusthe number of toys produced in a day. On aparticular day, the total cost of production wasRs 750. We would like to find out the number oftoys produced on that day.

Sol. (i) Let the number of marbles John had be x.

Number of marbles Jivanti had = 45 – x.The number of marbles left with John, when he lost

5 marbles = x – 5

The number of marbles left with Jivanti, when shelost 5 marbles = 45 – x – 5 = 40 – x

Then, required product = (x – 5) (40 – x)

= 40x – x2 – 200 + 5x

= –x2 + 45x – 200

As per condition : –x2 + 45x – 200 = 124

–x2 + 45x – 324 = 0

x2 – 45x + 324 = 0

x2 – 9x – 36x + 324 = 0

x(x – 9) – 36 (x – 9) = 0

(x – 36) (x – 9) = 0 x = 36 or x = 9

Hence, they had started with 36 and 9 marbles.

(ii) Let the number of toys produced on a particularday be x.

The cost of production of each toy on that day = Rs(55 – x)

The total cost of production on that day= Rs x(55 – x)

Therefore, x(55 – x) = 750

55x – x2 = 750

x2 – 55x + 750 = 0

x2 – 25x – 30x + 750 = 0 x(x – 25) – 30(x – 25) = 0

(x – 25)(x – 30) = 0

x = 25 or x = 30

Hence, the number of toys produced on a particularday can be 25 or 30.

Q.3. Find two numbers whose sum is 27 andproduct is 182.

Sol. Let first number = x.

And second number = 27 – x

Their product = x(27 – x)

As per condition : x(27 – x) = 182

27x – x2 = 182

–x2 + 27x – 182 = 0

x2 – 27x + 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) – 14(x – 13) = 0

(x – 13) (x – 14) = 0

x = 13 or x = 14

Hence, the required numbers are 13 and 14.

Q.4. Find two consecutive positive integers,sum of whose squares is 365.

Sol. Let first integer = x

Second consecutive integer = x + 1

Sum of their squares = x2 + (x + 1)2

= x2 + x2 + 1 + 2x = 2x2 + 2x + 1

As per condition : 2x2 + 2x + 1 = 365

2x2 + 2x = 364 x2 + x = 182

x2 + x – 182 = 0

x2 + 14x – 13x – 182 = 0

x(x + 14) – 13 (x + 14) = 0

(x + 14) (x – 13) = 0

x = 13 or x = –14

Since, x is a positive integer, therefore 14x ≠ − .

If x = 13 then, 2nd number = 13 + 1 = 14

Hence, the required two positive consecutiveintegers are 13 and 14.

Q.5. The altitude of a right triangle is 7 cmless than its base. If the hypotenuse is 13 cm,find the other two sides. [Imp.]

Sol. Let base of the right triangle = x cm

Height of the right triangle = (x – 7) cm

By Pythagoras theorem,

(Hyp.)2 = (Base)2 + (Height)2

(13)2 = x2 + (x – 7)2

169 = x2 + x2 + 49 – 14x

2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0

x2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

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(x – 12) (x + 5) = 0 x = 12 or x = –5

Side cannot be negative, so, x = 12 cmNow, x – 7 = 12 – 7 = 5 cm.Hence, the length of the other two sides are

5 cm and 12 cm.

Q.6. A cottage industry produces a certainnumber of pottery articles in a day. It wasobserved on a particular day that the cost ofproduction of each article (in rupees) was 3more than twice the number of articles producedon that day. If the total cost of production on thatday was Rs 90, find the number of articlesproduced and the cost of each article. [Imp.]

Sol. Let the number of articles produced = x

Cost of production of an article = Rs (2x + 3)

Total cost of production = Number of articlesproduced on that day × Cost of production of eacharticle

As per condition : x(2x + 3) = 90 2x2 + 3x – 90 = 0 2x2 + 15x – 12x – 90 = 0 x (2x + 15) –6(2x + 15) = 0

(2x + 15) (x – 6) = 0 x = 6 or x =15

2−

As, number of articles cannot be negative.

Therefore, x ≠ 15

2−

Hence, number of articles produced on that day= 6

So, cost of each article = 2x + 3= Rs (2 × 6 + 3)= Rs 15


Q.1. The roots of the quadratic equationx2 + 7x + 12 = 0 are : [2011 (T-II)]

(a) –4, –3 (b) 4, –3(c) 4, 3 (d) –4, 3

Sol. (a) x2 + 7x + 12 = 0

x2 + 3x + 4x + 12 = 0 x(x + 3) + 4(x + 3) = 0 (x + 3)(x + 4) = 0 x = –3, –4Q.2. The roots of the equation

x2 – 3 x – x + 3 = 0 are : [2011 (T-II)]

(a) 3 , 1 (b) – 3 , 1

(c) – 3 , –1 (d) 3 , –1

Sol. (a) x2 – 3 x – x + 3 = 0

x(x – 3 ) – 1(x – 3 ) = 0

(x – 3 )(x – 1) = 0 x = 3 , 1

Q.3. The roots of the quadratic equation

3 x2 – 2x – 3 = 0 are : [2011 (T-II)]

(a) – 3 ,1

3 (b) 2, 3

(c)3 2

, –2 3

(d) –13,

3

Sol. (d) 3 x2 – 2x – 3 = 0

3 x2 – 3x + x – 3 = 0

3 x(x – 3 ) + 1(x – 3 ) = 0

(x – 3 )( 3 x + 1) = 0 x = 3 ,–1

3

Q.4. The roots of 23 10 7 3 0p p+ + = are :

(a) 3 and 7 3 (b) 3 and7

3−

(c) 3− and7

3− (d) 3 and 7 3−

Sol. (c) We have, 23 10 7 3 0p p+ + =

23 3 7 7 3 0p p p+ + + =

3 ( 3) 7( 3) 0p p p+ + + =

( 3)( 3 7) 0p p+ + =

p = 3− or p =7

3

−

Q.5. The roots of x2 – x – a(a + 1) = 0 are :(a) a and (a – 1) (b) (a – 1) and a

(c) (a + 1) and (a – 1)

(d) –a and (a + 1)

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Sol. (d) We have, x2 – x – a(a + 1) = 0

x2 – (a + 1) x + ax – a (a + 1) = 0 x{x – (a + 1)} + a{x –(a + 1)} = 0 [x – (a + 1)](x + a) = 0 x = a + 1 or x = –a

Q.6. (2 + 2 3 ) and (2 – 2 3 ) are roots of :

(a) 23 2 2 2 3 0x x+ − =(b) x2 – 4x – 8 = 0(c) x2 – x + 1 = 0(d) none of these

Sol. (b) x2 – 4x – 8 = 0

2 2 2 3 2 2 3 x x x

2 2 3 2 2 3 0

2 2 3 2 2 3 x x

2 2 3 0 x

2 2 3 2 2 3 0 x x

x = 2 + 2 3 or x = 2 – 2 3

So, they are roots of x2 – 4x – 8 = 0

Q.7. The roots of a2x2 + (a2 – b2)x – b2 = 0are :

(a) 1 anda

b(b) –1 and

2

2

b

a

(c) ab and 1 (d) 1 and2

2

a

b

−

Sol. (b) We have, a2x2 + (a2 – b2)x – b2 = 0

a2x (x + 1) – b2(x + 1) = 0 (a2x – b2) (x + 1) = 0

x = –1 and x =

2

2

b

aQ.8. When k = 3, the roots of

(4 – k)x2 + (2k + 4)x + (8k + 1) = 0 are :

(a) 5, 5 (b) –5, 5 (c) 5,1

5(d) –5, –5

Sol. (d) When k = 3,

(4 – k)x2 + (2k + 4)x + (8k + 1) = 0 x2 + 10x + 25 = 0 (x + 5) (x + 5) = 0 x = –5, –5

Q.9. The roots of the equation3a2x2 + 8abx + 4b2 = 0 are :

(a)2

,b b

a a(b)

2,

3

b b

a a

(c)2

, b b

a a(d)

2 2,

3

b b

a a

Sol. (d) We have, 3a2x2 + 8abx + 4b2 = 0

3a2x2 + 6abx + 2abx + 4b2 = 0 3ax(ax + 2b) + 2b(ax + 2b) = 0 (3ax + 2b) (ax + 2b) = 0

x =2b

a

−or x =

2

3

b

a

−

Q.10. x =2

a band x =

2

a b−are roots of

the quadratic equation :

(a) 4x2 + 4a2x + (a4 – b4) = 0(b) 4x2 – 2(a2 + b2)x + a2b2 = 0(c) x2 – 2ax + a2 – b2 = 0(d) 4x2 – 4ax + (a2 – b2) = 0

Sol. (d) 4x2 – 4ax + (a2 – b2) = 04x2 – 2(a + b)x – 2(a – b)x

+ (a2 – b2) = 02x{2x – (a + b)} – (a – b)

{2x – (a + b)} = 0{2x – (a + b)}{2x – (a – b)}

x =2

a bor x =

2

a b

Q.11. Find the roots of the following

equation :1 1 11

;4 7 30x x

− =+ −

x ≠ – 4, 7

[2008, 2011 (T-II)]

Sol. We have,1 1 11

4 7 30x x− =

+ −

7 4 11

( 4)( 7) 30

x x

x x

− − −=

+ −

2

11

3 28x x

−− −

=11

30

x2 – 3x – 28 = –30 x2 – 3x + 2 = 0 x2 – 2x – x + 2 = 0 x(x – 2) – 1(x – 2) = 0 (x – 2) (x – 1) = 0 x = 2 or x = 1

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Q.12. Solve for x : abx2 + (b2 – ac)x – bc = 0[2005]

Sol. We have, abx2 + (b2 – ac)x – bc = 0

abx2 + b2x – acx – bc = 0 bx(ax + b) – c(ax + b) = 0 (ax + b) (bx – c) = 0ax + b = 0 or bx – c = 0 ax = – b or bx = c

x =b

a or x =

c

b

Q.13. Solve the following quadraticequations by factorisation method :

(i)1

1

x x

x x

=

34

15, x 0, x –1

[2011 (T-II)]

(ii)3 1

2

x x

x x

=

17

4, x 2, x 0

Sol. (i)1 34

1 15

x x

x x

( )2 2( 1) 34

1 15

x x

x x

+ +=

+

2 2

2

2 1 34

15

x x x

x x

+ + +=

+ 34x2 + 34x = 15x2 + 15x2 + 30x + 154x2 + 4x – 15 = 04x2 + 10x – 6x – 15 = 0 2x(2x + 5) – 3(2x + 5) = 0 (2x + 5) (2x – 3) = 0 2x + 5 = 0 or 2x – 3 = 0

x = –5

2or x =

3

2

(ii)3 1

2

x x

x x

=

17

4

3 1 2

2

x x x x

x x

17

4

( )+ − − − +

−

2 2

2

3 2 2

2

x x x x x

x x=

17

4

+

−

2

2

2 2

2

x

x x=

17

4

8x2 + 8 = 17x2 – 34x 9x2 – 34x – 8 = 0 9x2 – 36x + 2x – 8 = 0

9x(x – 4) + 2(x – 4) = 0 (x – 4) (9x + 2) = 0 x – 4 = 0 or 9x + 2 = 0

⇒ x = 4 or x = −2

9.

Q.14. Solve the following quadratic equationfor x :

p2x2 + (p2 – q2)x – q2 = 0 [2011 (T-II)]

Sol. We have, p2x2 + (p2 – q2)x – q2 = 0

p2x2 + p2x – q2x – q2 = 0

(p2x2 + p2x) – (q2x + q2) = 0

p2x(x + 1) – q2(x + 1) = 0

(x + 1)(p2x – q2) = 0 x + 1 = 0 or, p2x – q2 = 0

x = –1 or, x =

2

2

q

p

Q.15. Solve the following quadraticequations by factorisation method :

(i)4

3x

5

2 3x

,

−≠ 30,

2x

[2011 (T-II)]

(ii)

2 1 3 9

3 2 3 3 2 3

x x

x x x x= 0

[2011 (T-II)]

(iii) +− − − −

1 1

( 1)( 2) ( 2)( 3)x x x x

+ =− −

1 1

( 3)( 4) 6x x

Sol. (i)4

3x

5

2 3x

4 3x

x

5

2 3x

(4 – 3x) (2x + 3) = 5x

12 – x – 6x2 = 5x 6x2 + 6x – 12 = 0 x2 + x – 2 = 0 x2 + 2x – x – 2 = 0 x(x + 2) – 1(x + 2) = 0 (x + 2) (x – 1) = 0 x + 2 = 0 or x – 1 = 0

x = –2 or x = 1.(ii) Clearly, the given equation is valid if

x – 3 0 and 2x + 3 0, i.e.,

when x –3

2, 3

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Now,

2 1 3 9

3 2 3 3 2 3

x x

x x x x= 0

2x(2x + 3) + (x – 3) + 3x + 9 = 0

[Multiplying throughout by (x – 3)(2x + 3)]

4x2 + 6x + x – 3 + 3x + 9 = 0

4x2 + 10x + 6 = 0

2x2 + 5x + 3 = 0

2x2 + 2x + 3x + 3 = 0

2x(x + 1) + 3(x + 1) = 0

(x + 1) (2x + 3) = 0

x + 1 = 0 [ 2x + 3 0]

x = –1Hence, x = –1 is the only solution of the given

equation.

(iii) 1 1

( 1)( 2) ( 2)( 3)x x x x

1 1

( 3)( 4) 6x x

3 1 1 1

( 1)( 2)( 3) ( 3)( 4) 6

x x

x x x x x

2 4 1 1

( 1)( 2)( 3) ( 3)( 4) 6

x

x x x x x

2 1 1

( 1)( 3) ( 3)( 4) 6x x x x

− + −=

− − −2 8 1 1

( 1)( 3)( 4) 6

x x

x x x

−

=− − −

3 9 1

( 1)( 3)( 4) 6

x

x x x

=− −

3 1

( 1)( 4) 6x x

(x – 1)(x – 4) = 18 x2 – 5x + 4 = 18 x2 – 5x – 14 = 0 x2 – 7x + 2x – 14 = 0 x(x – 7) + 2(x – 7) = 0 (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2.

Q.16. Solve : = + ++ +

1 1 1 1,

a b x a b x

0,x ( )x a b . [2005, 2011 (T-II)]

Sol. We have,+ +

1

( )a b x= + +

1 1 1

a b x

−+ +

1 1

a b x x= +

1 1

a b

− + + +

=+ +

( )

( )

x a b x b a

x a b x ab

( )

( )− +

+ +a b

x a b x=

+a b

ab

( )−

+ +1

x a b x=

1

ab

[On dividing both sides by (a + b)] x(a + b + x) = – ab x2 + ax + bx + ab = 0 x(x + a) + b(x + a) = 0 (x + a) (x + b) = 0 x + a = 0 or x + b = 0

x = – a or x = – bHence, –a and –b are the roots of the given

equation.

Q.17. Solve : + =− − −

2a b c

x a x b x c

[Imp.]

Sol. We have + =− − −

2a b c

x a x b x c

− + − =− − −

( ) ( ) 2

( )( )

a x b b x a c

x a x b x c

− + −

=−− + +2

2

( )

ax ab bx ab c

x cx a b x ab

+ −

=−− + +2

( ) 2 2

( )

a b x ab c

x cx a b x ab

(a + b)x2 – 2abx – c(a + b)x + 2abc= 2cx2 – 2c(a + b)x + 2abc

x2(a + b – 2c) – x(2ab + ac + bc – 2ac– 2bc) = 0

x2(a + b – 2c) – x(2ab – ac – bc) = 0 x{x(a + b – 2c) – (2ab – ac – bc)} = 0

x = 0 or x =− −

+ −2

2

ab ac bc

a b c.

Q.18. Solve : ++ + +

2 a a bx

a b ax + 1 = 0

0,x ( )x a b . [2005]

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Sol. We have, +

+ + + + 2 1

a a bx x

a b a= 0

+ +

+ + + ×+ +

2 ax a b a a bx x

a b a a b a= 0

++ + + + +

a a b ax x x

a b a a b= 0

+ + + +

a a bx x

a b a= 0

++a

xa b

= 0 or+

+a b

xa

= 0

−+a

xa b

= or+

= −a b

xa

.

Q.19. Solve : 4x2 – 4a2 x + (a4 – b4) = 0[2004]

Sol. We have, 4x2 – 4a2x + (a4 – b4) = 0

Here, constant term= a4 – b4 = (a2 – b2) (a2 + b2) and,

coefficient of middle term = – 4a2

= – {2(a2 + b2) + 2(a2 – b2)} 4x2 – 4a2x + (a4 – b4) = 0

4x2 – {2(a2 + b2) + 2(a2 – b2)}x+ (a2 – b2)(a2 + b2) = 0

4x2 – 2(a2 + b2)x – 2(a2 – b2)x

+ (a2 – b2)(a2 + b2) = 0 {4x2 – 2(a2 + b2)x} – {2(a2 – b2)x

– (a2 – b2)(a2 + b2)} = 0

2x {2x – (a2 + b2)} – [{(a2 – b2){2x – (a2 + b2)}] = 0

{2x – (a2 + b2)}{2x – (a2 – b2)} = 0

2x – (a2 + b2) = 0 or 2x – (a2 – b2) = 0

x =2 2

2

a b+or x =

2 2

2

a b−.

Q.20. Solve :9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0

[2004, 2009, 2011 (T-II)]Sol. Here, constant term = 2a2 + 5ab + 2b2

= 2a2 + 4ab + ab + 2b2

= 2a(a + 2b) + b(a + 2b)

= (a + 2b)(2a + b)

and, coefficient of middle term = –9(a + b)

= –3{(2a + b) + (a + 2b)}

9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0

9x2 – 3{(2a + b) + (a + 2b)}x

+ (2a + b)(a + 2b) = 0

9x2 – 3(2a + b)x – 3(a + 2b)x

+ (2a + b)(a + 2b) = 0

3x{3x – (2a + b)} – (a + 2b){3x

– (2a + b)} = 0

{3x – (2a + b)} {3x – (a + 2b)} = 0

{3x – (2a + b)} = 0 or {3x – (a + 2b)} = 0

+

=2

3

a bx or

+=

2

3

a bx .

Q.21. Solve : + −+( 1) (1 )4 4x x = 10 [2004]

Sol. We have, + −+( 1) (1 )4 4x x = 10

4x.4 + 4.4–x = 10

4x.4 +4

4x= 10

4y +4

y= 10, where 4x = y

4y2 + 4 = 10y 4y2 – 10y + 4 = 0 4y2 – 8y – 2y + 4 = 0 4y(y – 2) – 2(y – 2) = 0 (y – 2) (4y – 2) = 0 y – 2 = 0 or 4y – 2 = 0

y = 2 or y =2

4=

1

2

4x = 2 or 4x =1

2[ 4x = y]

22x = 21 or 22x = 2–1

2x = 1 or 2x = –1

x =1

2or x = –

1

2.

Q.22. Solve for x :12abx2 – (9a2 – 8b2)x – 6ab = 0 [2006]Sol. We have, 12abx2 – (9a2 – 8b2)x – 6ab = 0

12abx2 – 9a2x + 8b2x – 6ab = 03ax(4bx – 3a) + 2b(4bx – 3a) = 0 (4bx – 3a)(3ax + 2b) = 0 4bx – 3a = 0 or 3ax + 2b = 0 4bx = 3a or 3ax = –2b

x =3

4

a

bor x =

2

3

b

aQ.23. Solve for x : 6x2 – 12ax + (a2 – b2) = 0

[2011 (T-II)]Sol. We have, 36x2 – 12ax + (a2 – b2) = 0

(36x2 – 12ax + a2) – b2 = 0

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(6x – a)2 – b2 = 0 (6x – a + b)(6x – a – b) = 0

6x – a + b = 0 or 6x – a – b = 0 6x = (a – b) or 6x = (a + b)

x =–

6

a bor x =

6

a b

Q.24. Solve for x :

1 3

2 4

− −+− −

x x

x x=

13

3(x 2, 4) [2005]

Sol. We have,1 3 1

32 4 3

x x

x x

( 1)( 4) ( 2)( 3)

( 2)( 4)

x x x x

x x=

10

3

3(x2 – 4x – x + 4 + x2 – 3x – 2x + 6)= 10(x2 – 4x – 2x + 8)

3(2x2 – 10x + 10) = 10 (x2 – 6x + 8)

6x2 – 30x + 30 = 10x2 – 60x + 80 10x2 – 60x + 80 – 6x2 + 30x – 30 = 0 4x2 – 30x + 50 = 0

2x2 – 15x + 25 = 0 2x2 – 5x – 10x + 25 = 0 x(2x – 5) – 5(2x – 5) = 0

(2x – 5) (x – 5) = 0

x =5

2or x = 5.

Q.25. Solve for x :

4 3 2 110 3

2 1 4 3

x x;

x x

− + − = + −

1;

2x

3

4x ≠

[2004, 2011 (T-II)]

Sol. We have,4 3 2 1

10 32 1 4 3

x x

x x

− + − = + −

Let4 3

2 1

x

x

−+

= y

Putting this value in the equation, we get

103 y

yy2 – 10 = 3y

y2 – 3y – 10 = 0 y2 – 5y + 2y – 10= 0

y(y – 5) + 2(y – 5) = 0 (y – 5) (y + 2) = 0 y = 5 or y = –2

When y = 5, When y = 2,

4 3 5

2 1 1

x

x

−=

+4 3 2

2 1 1

x

x

10x + 5 = 4x – 3 – 4x – 2 = 4x – 3 6x = –8 – 4x – 4x = –3 + 2

x =4

3 8x = 1 x =

1

8

Q.26. Solve for x :

+ − − = − +2 3 3

2 25 5;3 2 3

x x

x xgiven that

x ≠ 3, x ≠3

.2

[2004]

Sol. We have,2 3 3

2 25 53 2 3

x x

x x

+ − − = − +

25

2 5yy

− =2 3

Let3

xy

x

+ = − 2y2 – 25 = 5y 2y2 – 5y – 25 = 0 2y2 – 10y + 5y – 25 = 0 2y(y – 5) + 5(y – 5) = 0 (y – 5) (2y + 5) = 0

y = 5 or y =5

2

−

When y = 5, When y =5

2

−

2 3

3

x

x

+−

= 52 3

3

x

x

+−

=5

2

−

5x – 15 = 2x + 3 4x + 6 = – 5x + 15 5x – 2x = 3 + 15 9x = 9 x = 1 3x = 18 x = 6

Solve the following quadratic equations byfactorisation :

1. 2 ( 2 1) 2 0x x− + + =

2. 2 ( 3 1) 3 0x x− + + =

3. 4x2 + 4bx – (a2 – b2) = 0

4. ax2 + (4a2 – 3b)x – 12ab = 0

5. a2x2 + (a2 + b2)x + b2 = 0, 0a ≠

6.2 1 2

m

m nx x

n

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7. (a + b)2 x2 – 4abx – (a – b)2 = 0

8. a(x2 + 1) – x(a2 + 1) = 09. x2 – x – a(a + 1) = 0

10. (x – 3) (x – 4) = 2

34

(33)

11.24 6 13 2 6 0 x x

12.1 2 6

, ( 0)2 1

xx x x

[Imp.]

13.1

,( 2)1

21

22

y y

y

= ≠−

−−

[Imp.]

14.1 1

a ba b

ax bx+ = +

− −[Imp.]

15.x a x b a b

x b x a b a

[Imp.]

16.2

5 6 0, ( 1)1 1

x xx

x x

17. 5(x + 1) + 5(2 – x) = 53 + 118. 22x – 3.2x + 2 + 32 = 0

19. 2 11 0x a x

a

+ + + =

20.2 22

0a c

ad x x c bb d

+ + =

21. 24 3 5 2 3 0x x+ − = [2006C]

22. a2b2x2 + b2x – a2x – 1 = 0

[2005, 2011 (T-II)]

23.2 1 3

2 3 5,3 2 1

x x

x x

− + − = + −given that

x ≠ 3, x ≠ 1.

2[2004]

24.1 3

2 7 5,3 1

x x

x x

− + − = + −given that

3, 1.x x≠ − ≠ [2004]25. 4x2 – 2(a2 + b2)x + a2b2 = 0 [2004]

26. 27 6 13 7 0y y− − = [2004, 2011 (T-II)]

4.3. SOLUTION OF A QUADRATICEQUATION BY COMPLETING THESQUARE : FORMULA METHOD

1. The roots of a quadratic equation can alsobe found by using the method of completing thesquare.

2. The roots of the quadratic equationax2 + bx + c = 0, a, b, c R and a 0 are given

by x = 2 4

2

b b ac

a

(Shridharacharya's formula)

TEXTBOOK'S EXERCISE 4.3

Q.1. Find the roots of the followingquadratic equations, if they exist, by the methodof completing the square :

(i) 2x2 – 7x + 3 = 0(ii) 2x2 + x – 4 = 0

(iii) 24 4 3 3 0x x+ + = [2011 (T-II)]

(iv) 2x2 + x + 4 = 0

Sol. (i) Given equation is : 2x2 – 7x + 3 = 0

2x2 – 7x = –3

Dividing both sides by 2, we get 2 7 3

2 2x x

−− =

Adding2

7

4

− to both sides, we get

2 22 7 7 3 7

2 4 2 4x x

− − − − + = +

2

7 3 49

4 2 16x

− − = +

2

7 24 49

4 16x

− + − =

2

7 25

4 16x

− =

7 25 5

4 16 4x − = ± = ±

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Case I : When,7 5

4 4x − =

Then,5 7 5 7

4 4 4x

+= + =

12

4x = = 3

Case II : When,7 5

4 4x

−− =

Then,5 7 5 7

4 4 4x

− − += + =

2 1

4 2x = =

Hence, roots of quadratic equation are1

2and 3.

(ii) Given equation is : 2x2 + x – 4 = 0 2x2 + x = 4

Dividing both sides by 2, we get 2 1 4

2 2x x+ =

Adding2

1

4

to both sides, we get

22 1 1

2 4x x

+ +

21

24

= +

21 1

24 16

x + = +

21 32 1

4 16x

+ + =

21 33

4 16x

+ =

1 33

4 16x

1 33

4 4x

+ = ±

Case I : When,1 33

4 4x + =

Then,33 1

4 4x

33 1

4x

−=

Case II : When1 33

4 4x

−+ =

Then,33 1

4 4x

−= − x =33 1

4

Hence, roots of quadratic equation are

1 33

4

− +and

1 33

4

− −.

(iii) Given equation is : 24 4 3 3 0x x+ + = 24 4 3 3x x+ = −

Dividing both sides by 4, we get 2 4 3 3

4 4x x

−+ =

2 3

34

x x−

+ =

Adding

23

2


22 3

32

x x

+ + =

23 3

4 2

−+

23 3 3

2 4 4x

−+ = +

23

02

x

+ =

3 3

02 2

x x

+ + =

3

2x

or

3

2x

−=

Hence, roots of given quadratic equation are

3

2

−and

3

2

−.

(iv) Given equation is : 2x2 + x + 4 = 0 2x2 + x = – 4

Dividing both sides by 2, we get2 1 4

2 2x x

−+ =

Adding2

1

4


22 1 1

2 4x x

+ + =

21

24

− +

2

1 12

4 16x

+ = − +

2

1 32 1

4 16x

− + + =

21 31

04 16

x− + = <

Since, square of any number cannot be negative.

So,2

1

4x

+ cannot be negative for any real x.

There is no real x which satisfies the givenquadratic equation. Hence, given quadratic equationhas no real roots.

Q.2. Find the roots of the quadraticequations given in Q.1 above by applying thequadratic formula.

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Sol. (i) Given equation is : 2x2 – 7x + 3 = 0

This is of the form ax2 + bx + c = 0

Where a = 2, b = –7, c = 3

Now, b2 – 4ac = (–7)2 – 4 × 2 × 3 = 49 – 24 = 25

x =2 4

2

b b ac

a

=( 7) 25 7 5

2 2 4

=7 5

4

+and

7 5

4

−

=12

4and

2

4= 3 and

1

2

Hence, 3 and1

2are the roots of given quadratic

equation.

(ii) Given equation is : 2x2 + x – 4 = 0

This is of the form ax2 + bx + c = 0.

Where a = 2, b = 1, c = – 4

Now, b2 – 4ac = (1)2 – 4 × 2 × (– 4) = 1 + 32 = 33

x =2 4

2

b b ac

a

− ± −

=1 33 1 33

2 2 4

− ± − ±=×

Hence,1 33

4

− +and

1 33

4

− −are the roots of

given quadratic equation.

(iii) Given equation is : 24 4 3 3 0x x+ + =This is of the form ax2 + bx + c = 0

Where a = 4, b = 4 3 , c = 3

Now, b2 – 4ac = 2(4 3) – 4 × 4 × 3

= 48 – 48 = 0

x =2 4

2

b b ac

a

=

4 3 0

2 4

=4 3 4 3 3 3

and and8 8 2 2

Hence,3

2

and

3

2

are the roots of given

quadratic equation.

(iv) Given equation is : 2x2 + x + 4 = 0This is of the form ax2 + bx + c = 0.Where, a = 2, b = 1, c = 4Now, b2 – 4ac = (1)2 – 4 × 2 × 4 = 1 – 32

= –31 < 0As, b2 – 4ac < 0, therefore the given quadratic

equation will have no real roots.

Q.3. Find the roots of the followingequations :

(i)1

3 0 x , xx

(ii)1 1 11

4 7 30,

x x

4,7x [2011 (T-II)]

Sol. (i) We have,1

3xx

− = 2 1

3x

x

−=

x2 – 1 = 3xx2 – 3x – 1 = 0This is of the form ax2 + bx + c = 0.Where a = 1, b = – 3, c = –1Now, b2 – 4ac = (–3)2 – 4 × 1 × (–1) = 9 + 4 = 13

x =2 4

2

b b ac

a

− ± −

=( 3) 13 3 13

2 1 2

Hence,3 13

2

and

3 13

2

−are the roots of given

quadratic equation.

(ii)1 1

4 7x x

=

11

30

( 7) ( 4) 11

( 4)( 7) 30

x x

x x

2

7 4 11

307 4 28

x x

x x x

2

11 11

303 28x x

– 11 × 30 = 11(x2 – 3x – 28) –30 = x2 – 3x – 28 x2 – 3x – 28 + 30 = 0 x2 – 3x + 2 = 0This is of the form ax2 + bx + c = 0.Where, a = 1, b = –3, c = 2Now, b2 – 4ac = (–3)2 – 4 × 1 × 2 = 9 – 8 = 1

x =2 4

2

b b ac

a

− ± −

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=( 3) 1 3 1

2 1 2

So, roots are3 1

2

+and

3 1

2

−

or4

2and

2

2or 2 and 1

Hence, 2 and 1 are the roots of given quadraticequation.

Q.4. The sum of the reciprocals of Rehman'sages (in years) 3 years ago and 5 years from now

is1

.3

Find his present age.

Sol. Let Rehman's present age = x years

3 years ago Rehman's age = (x – 3) years5 years from now Rehman's age = (x + 5) years

As per condition :1 1 1

3 5 3x x+ =

− +

5 3 1

( 3)( 5) 3

x x

x x

+ + −=

− + 2

2 2 1

35 3 15

x

x x x

+=

+ − −

2

2 2 1

32 15

x

x x

+=

+ − 6x + 6 = x2 + 2x – 15

x2 + 2x – 15 – 6x – 6 = 0

x2 – 4x – 21 = 0,

which is a quadratic equation in x. This is of theform ax2 + bx + c = 0

Where, a = 1, b = – 4, c = –21

Now, b2 – 4ac = (– 4)2 – 4 × 1 × (–21)

= 16 + 84 = 100

So, x =2 4

2

b b ac

a

( 4) 100 4 10

2 1 2

x

So, roots are4 10

2

+and

4 10

2

−or 7 and –3

Age cannot be negative.∴ x = 7Hence, Rehman's present age = 7 years.

Q.5. In a class test, the sum of Shefali'smarks in Mathematics and English is 30. Hadshe got 2 marks more in Mathematics and3 marks less in English, the product of theirmarks would have been 210. Find her marks inthe two subjects.

Sol. Let Shefali got x marks in Mathematics.

∴Shefali's marks in English = 30 – xAs per condition :

Shefali's marks in Mathematics = x + 2

Shefali's marks in English = 27 – x

∴Required product = (x + 2) (27 – x)= 27x – x2 + 54 – 2x

= –x2 + 25x + 54

According to the given problem,

–x2 + 25x + 54 = 210

–x2 + 25x + 54 – 210 = 0

–x2 + 25x – 156 = 0

x2 – 25x + 156 = 0

x2 – 12x – 13x + 156 = 0

(x – 12) (x – 13) = 0 x = 12 or x = 13

Case I : When x = 13, then

Shefali's marks in Mathematics = 13

Shefali's marks in English = 30 – 13 = 17

Case II : When x = 12, then

Shefali's marks in Mathematics = 12

Shefali's marks in English = 30 – 12 = 18

Hence, Shefali's marks in the two subjects are 13and 17 or 12 and 18.

Q.6. The diagonal of a rectangular field is60 metres more than the shorter side. If thelonger side is 30 metres more than the shorterside, find the sides of the field.

Sol. Let shorter side of the rectangular field

= AD = x mC

B

D

A

(+ 60 m)

x

Longer side of the rectangular field = AB= (x + 30)m

Diagonal of rectangular field = DB = (x + 60) m

Using Pythagoras Theorem in DAB,

we have, (DB)2 = (AD)2 + (AB)2

(x + 60)2 = (x)2 + (x + 30)2

x2 + 3600 + 120x = x2 + x2 + 900 + 60x

x2 + 3600 + 120x – 2x2 – 900 – 60x = 0

–x2 + 60x + 2700 = 0

x2 – 60x – 2700 = 0

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Using the quadratic formula, we get

x =2( 60) ( 60) 4 1 2700

2 1

60 14400

2

=

60 120

2

±

So, roots are60 120

2

+and

60 120

2

−

or180

2and

60

2

−or 90 and –30

A side cannot be negative, so we will take thepositive value.

So, x = 90Hence, shorter side of rectangular field = 90 m∴ Longer side of rectangular field

= (90 + 30) m = 120 m

Q.7. The difference of squares of twonumbers is 180. The square of the smallernumber is 8 times the larger number. Find thetwo numbers. [2011 (T-II)]

Sol. Let larger number = x. Smaller number = y

As per condition :x2 – y2 = 180 …(i) ( x > y)Also, y2 = 8x …(ii)From eqn. (i) and (ii), we have x2 – 8x = 180 x2 – 8x – 180 = 0Using the quadratic formula, we get

x =2( 8) ( 8) 4 1 ( 180)

2 1

=8 784 8 28

2 2

So, roots are8 28

2

+and

8 28

2

−

or36

2and

20

2

−or 18 and –10.

When, x = –10, then from eqn. (ii),

y2 = 8(–10)

= – 80, which is impossible.

When, x = 18, then from eqn. (ii),

y2 = 8(18) = 144

144y = ± 12y = ±Hence, required numbers are 18 and 12 or 18

and –12.

Q.8. A train travels 360 km at a uniformspeed. If the speed had been 5 km/h more, itwould have taken 1 hour less for the samejourney. Find the speed of the train.

[2011 (T-II)]Sol. Let speed of the train = x km/hour

Distance covered by the train = 360 km

Time =Distance

Speed=

360

xhours

Increased speed of the train = (x + 5) km/hour∴ Time taken by the train with increased speed

=360

5x hours

As per condition :360 360

15x x

− =+

360( 5) 360

( 5)

x x

x x

+ −+ = 1

2

360 1800 3601

5

x x

x x

+ −=

+ 1800 = x2 + 5x

x2 + 5x – 1800 = 0


x

25 5 4 1( 1800)

2 1

=5 7225 5 85

2 2

So, roots are5 85

2

and

5 85

2

or80

2and

90

2

or 40 and – 45.

Since, speed cannot be negative. So, we will takethe positive value.

So, x = 40.

Hence, speed of train = 40 km/hour.

Q.9. Two water taps together can fill a tank

in3

98

hours. The tap of larger diameter takes 10

hours less than the smaller one to fill the tankseparately. Find the time in which each tap canseparately fill the tank. [2008, 2011 (T-II)]

Sol. Let time taken by larger tap = x hours

Time taken by smaller tap = (x + 10) hours

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Portion of tank filled in one hour by the larger

tap =1

xPortion of tank filled in one hour by the smaller

tap =1

10x + Portion of tank filled in one hour by both taps

=1

x+

1

10x +…(i)

But, two taps together can fill the tank in3

98

hours, i.e., in75

8hours.

Now portion of tank filled by two taps together in

one hour =8

75…(ii)

From eqn. (i) and eqn. (ii), we have

1 1 8

10 75x x+ =

+

10

( 10)

x x

x x

+ ++ =

8

75⇒ 2

2 10 8

7510

x

x x

+ =+

75(2x + 10)= 8 (x2 + 10x)

150x + 750 = 8x2 + 80x

8x2 + 80x – 150x – 750 = 0

8x2 – 70x – 750 = 0

4x2 – 35x – 375 = 0Using the quadratic formula, we get

2( 35) ( 35) 4 4 ( 375)

2 4x

35 7225 35 85

8 8

So, roots are35 85

8

+and

35 85

8

−.

or120

8and

50

8

−or 15 and

25

4

−

Since x is the time, it cannot be negative.Therefore, x = 15.

Hence, larger water tap will fill the tankseparately in 15 hours.

Smaller water tap will fill the tank separately in(15 + 10) = 25 hours.

Q.10. An express train takes 1 hour less thana passenger train to travel 132 km betweenMysore and Bangalore (without taking into

consideration the time they stop at intermediatestations). If the average speed of the expresstrain is 11 km/hour more than that of thepassenger train, find the average speed of thetwo trains. [2011 (T-II)]

Sol. Let speed of passenger train = x km/hour

So, speed of express train = (x + 11) km/hourDistance between Mysore and Bangalore

= 132 km

Time taken by passenger train =132

hourx

Time taken by express train =132

hour11x +

As per condition :132 132

111x x

− =+

132( 11) 132

1( 11)

x x

x x

+ − =+

2

132 1452 1321

11

x x

x x

+ − =+

1452 =2 11x x+

x2 + 11x – 1452 = 0Using the quadratic formula, we get

211 11 4 1 ( 1452)

2 1

x

=11 5929 11 77

2 2

So, roots of the equation are

–11 7711 77 66 88and or and

2 2 2 2

or 33 and – 44.Since, x is the speed of the train, it cannot be

negative.Hence, speed of passenger train = 33 km/hourSpeed of express train = (33 +11) = 44 km/hour.

Q.11. Sum of the areas of two squares is 468m2. If the difference of their perimeters is24 m, find the sides of the two squares.

[2011 (T-II)]Sol. Let length of side of one square = x m

Area of square = 2 2mx

Perimeter of square = 4x mLet length of side of another square = y m

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Area of square = 2 2my

Perimeter of square = 4y mAs per condition,

2 2 468x y+ = ...(i)

And 4 4 24x y− =

4( ) 24x y− =

6x y− = ...(ii)

From (i) and (ii), we have 2 2(6 ) 468 y y

2 236 12 468 y y y

22 12 36 468 0 y y

22 12 432 0 y y

2 6 216 0 y y


26 6 4 1 ( 216)

2 1

y

6 900 6 30

2 2

–6 –306 30 24 –36and and

2 2 2 2y y

12 and – 18

As, length cannot be negative. So, we will takethe positive value.

So, y = 12

From (ii), x = 6 + 12 = 18

Hence, sides of two squares are 12 m and 18 m.


Q.1. Find the roots of the following equation4x2 + 4bx – (a2 – b2) = 0 by the method ofcompleting the square.

Sol. We have, 4x2 + 4bx – (a2 – b2) = 0

x2 + bx – −

22

4

a b = 0

x2 + 2 2

bx =

−2 2

4

a b

x2 + 2 2

bx +

2

2

b=

−2 2

4

a b+

2

2

b

+

2

2

bx =

2

4

ax +

2

b= ±

2

a

x = –2

b±

2

a x =

− −2

b a,

− +2

b a

Hence, the roots are : – and2 2

a b a b+ −

.

Q.2. Solve 2x2 + 14x + 9 = 0, by completingthe squares, when (i) x is a rational number,(ii) x is a real number.

Sol. Given equation can be rewritten as2x2 + 14x = –9

x2 + 7x = −9

2

x2 + 7x +

27

2=

−

27 9

2 2

+

27

2x = −

49 9

4 2

+

27

2x =

−=

49 18 31

4 4

+

27

2x =

±

231

4 +

7

2x = ± 31

2

x = − ±7 31

2 2

7 31

2

The roots are− +7 31

2and

− −7 31

2,

which are both irrational.Thus, (i) when x is a rational number, the

equation has no solution.(ii) when x is a real number, the roots are

7 31

2

− +and

7 31

2

− −

Q.3. Using the quadratic formula, solve theequation :

2 2 2 4 4 2 2(4 3 ) 12 0a b x b a x a b [2006]

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x =+ + −2 21 1

2

a a

aor

+ − +2 21 1=

2

a ax

a

Hence, x = a,1

a.

(ii) We have, x2 + 2ab = (2a + b)x x2 – (2a + b)x + 2ab = 0Here, A = 1, B = – (2a + b), C = 2abB2 – 4AC = [–(2a + b)]2 – 4 × 1 × 2ab= (2a + b)2 – 8ab= 4a2 + b2 + 4ab – 8ab= 4a2 + b2 – 4ab = (2a – b)2

Using quadratic formula, we get

x =2B B 4AC

2A

x( ) ( )+ ± −

=×

22 2

2 1

a b a b

( )+ ± −=

2 (2 )

2

a b a b

+ + −

=2 2

2

a b a bx

or+ − +

=2 2

2

a b a bx

x =4

2

a= 2a or x =

2

2

b= b

Hence, x = 2a, b.

(iii) We have, 3y2 + (6 + 4a)y + 8a = 0

Here, A = 3, B = 6 + 4a, C = 8a

B2 – 4AC = (6 + 4a)2 – 4 × 3 × 8a

= 36 + 16a2 + 48a – 96a

= 36 + 16a2 – 48a = (6 – 4a)2

Using quadratic formula, we get

y =2B B 4AC

2A

y =( ) ( )− + ± −

×

26 4 6 4

2 3

a a

( ) ( )− + ± −=

6 4 6 4

6

a a

− − + −

=6 4 6 4

6

a ay

or6 4 6 4

6

a a

Sol. We have, a2b2x2 – (4b4 – 3a4)x – 12a2b2 = 0

Here, A = a2b2, B = –(4b4 – 3a4), C = –12a2b2

D =2B 4AC

4 4 2 2 2 2 2[ (4 3 )] 4 ( 12 )b a a b a b= − − − −

=8 8 4 4 4 416 9 24 48b a a b a b+ − +

= 16b8 + 9a8 + 24a4b4

=4 2 4 2 4 4(4 ) (3 ) 2(4 )(3 ) b a b a

= 4 4 2(4 3 )b a [ 2 2 22 ( )x y xy x y+ + = + ]

4 4D (4 3 )b a= +

Now,B D

2Ax

4 4 4 4

2 2

[ (4 3 ) (4 3 )

2

b a b ax

a b

− − − ± +=

4 4 4 4

2 2

4 3 4 3

2

b a b ax

a b

− + +=

Or,

4 4 4 4

2 2

4 3 4 3

2

b a b ax

a b

− − −=

4 2

2 2 2

8 4

2

b bx

a b a= = or,

4 2

2 2 2

6 3

2

a ax

a b b

− −= =

Q.4. Solve each of the following equationsby using quadratic formula :

(i) a(x2 + 1) = x(a2 + 1)

(ii) x2 + 2ab = (2a + b)x

(iii) 3y2 + (6 + 4a)y + 8a = 0

(iv) (x + 4)(x + 5) = 3(x + 1)(x + 2) + 2x

Sol. (i) We have, a(x2 + 1) = x(a2 + 1)

ax2 – x(a2 + 1) + a = 0, which is quadratic in x.Here, A = a, B = –(a2 + 1) and C = aUsing quadratic formula, we get

x =2A B 4AC

2A

x( ) ( )+ ± + −

=

22 2 21 1 4

2

a a a

a

( ) ( )+ ± −=

2 21 1

2

a a

a

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8 12

or =6 6

ay y

4

= – or = –23

ay y .

(iv) (x + 4)(x + 5) = 3(x + 1)(x + 2) + 2x

x2 + 9x + 20 = 3(x2 + 3x + 2) + 2x x2 + 9x + 20 = 3x2 + 11x + 6 2x2 + 2x – 14 = 0 x2 + x – 7 = 0Here, a = 1, b = 1, c = –7

x =− ± −2 4

2

b b ac

a

=− ± − −

×

21 1 4(1)( 7)

2 1

x =− ± + − ±

=1 1 28 1 29

2 2

x =− + − −1 29 1 29

or2 2

.

Q.5. Using quadratic formula, solve thefollowing quadratic equation for x :

2 2 24 4 0x ax a b− + − = [2004]

Sol. Comparing it with 2A B C 0x x

A = 1, B = –4a, C = 4a2 – b2

2 2 2 2D = B 4AC ( 4 ) 4 1 (4 )a a b

=2 2 2 216 16 4 4a a b b− + =

2( 4 ) 4B D

2A 2 1

a bx

=4 2

22

a ba b

2 or = 2 x a b x a b

Q.6. Using quadratic formula, solve for x :29 3( ) 0x a b x ab− + + = [2005, 2011 (T-II)]

Sol. On comparing it with2A B C 0 x x

A 9, B 3( ), Ca b ab

D 2 2= B 4AC [ 3( )] 4 9 a b ab

=2 29( 2 ) 36a b ab ab

=2 29 9 18 36a b ab ab+ + −

= 2 29 9 18a b ab+ −

= 2 29( 2 )a b ab+ − 2 29( ) [3( )]a b a b= − = −

x=2[ 3( )] [3( )]B D

2A 2 9

a b a b

=3( ) 3( )

18

a b a b+ ± −

x = or6 6

+ + − + − +=

a b a b a b a bx

x 2 2or

6 6

a bx ⇒ or

3 3

a bx x= = .

Q.7. Using quadratic formula, solve the

following for x : 2 2 2 2 29 3( ) 0x a b x a b

[2005]

Sol. We have, 2 2 2 2 29 3( ) 0x a b x a b− + + =

On comparing it with2A B C 0x x , we get

A 9 , 2 2 2 2B 3( ) and Ca b a b

D = 2 2 2 2 2 2B 4AC 9( ) 4(9) ( )a b a b

= 4 4 2 2 2 29( 2 ) 36a b a b a b+ + −

=4 4 2 2 2 29( 2 4 )a b a b a b+ + −

= 4 4 2 2 2 2 29( 2 ) 9( )a b a b a b+ − = −

2 2 2 2 2[ 3( )] 9( )B D

2A 2 9

a b a bx

=2 2 2 23( ) 3( )

18

a b a b+ ± −

2 2 2 23 3 3 3

18

a b a bx

+ + −=

2 26

18 3

a ax = =

or2 2 2 23 3 3 3

18

a b a bx

+ − +=

2 26

18 3

b bx = =

Q.8. Using quadratic formula, solve the

following for x : 2 2( ) 0abx b ac x bc [2006]

Sol. We have,2 2( ) 0abx b ac x bc+ − − =

On comparing it with 2A B C 0x x , we get :

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Solve the following quadratic equations bycompleting the squares :

1. 22 3 2 2 0x x

2. 23 10 7 3 0 x x

3.2 ( 2 1) 2 0 x x

4. 2 4 2 6 0 x x

5. 2 2 23 2 0 a x abx b

6.2 1

1 02

y y

7. (5 2 )(3 4 ) 8 z a z b ab

Solve the following equations by using theformula :

8. 2 8 12 0 x x

9. 22 4 1 0 x x

10. 24 2 1 0 x x

11.23 2 1 0 x x

12. 2 27 6x ax a− = −

13. 27 6 13 7 0 x x

14. 23 (6 4 ) 8 0y a y a

15.2 2 25 4 0 a x abx b

16. 2 2 23 2 0 p q pqx x

17.20.6 0.1 1.5 0 x x

18.2 5 5 0 x x

19.2 2 3 6 0x ax x a

20. 2(2 ) 2 p q x x pq

21. 2 2 2 2 2 1 0 a b x b x a x

22. 2 2 24 4 a b ax x

23.2( ) ( ) ( ) 0 a b x b c x c a

24.2 22( ) ( ) 1 x m n x m n

A = ab, B = b2 – ac, C = –bc

D = 2 2 2B 4AC ( ) 4 ( )b ac ab bc

= 4 2 2 2 22 4b a c acb acb+ − +

= 4 2 2 2 2 22 ( )b a c acb b ac+ + = +

Using quadratic formula,B D

2Ax

2 2 2( ) ( )

2

b ac b ac

ab

− − ± +=

=2 2( ) ( )

2

b ac b ac

ab

− + ± +

2 2

2

b ac b acx

ab

− + + += =

2

2

ac c

ab b=

or2 2

2

b ac b acx

ab

− + − −= =

22

2

b b

ab a

Q.9. Solve for x : + =+ + +1 2 4

,1 2 4x x x

≠ ,1x − , −2 4 [2011 (T-II)]

Sol. We have, + =+ + +1 2 4

1 2 4x x x

+ = ++ + + +1 2 1 3

1 2 4 4x x x x

−+ +1 1

1 4x x= −

+ +3 2

4 2x x

+ − −+ +

4 1

( 1)( 4)

x x

x x=

+ − −+ +

3 6 2 8

( 4)( 2)

x x

x x

+ +

3

( 1) ( 4)x x=

−+ +

2

( 4)( 2)

x

x x

+3

1x=

−+

2

2

x

x

3x + 6 = x2 – x – 2

x2 – 4x – 8 = 0Here, a = 1, b = – 4 and c = – 8

x =− ± −2 4

2

b b ac

a

x± + ±

= =4 16 32 4 4 3

2 2= ±2 2 3

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4.4. NATURE OF ROOTS

1. The roots of the quadratic equation,

2 0, 0 ax bx c a are given by

2 4

2

b b acx

a

The expression 2 4b ac is called the

discriminant of the quadratic equation2 0ax bx c .

2. The discriminant, usually denoted by D,decides the nature of roots of a quadraticequation.

(i) If D > 0, the equation has real roots andare unequal, i.e., unequal-real roots.

If D is a perfect square, the equation hasunequal-rational roots.

(ii) If D = 0, the equation has real and equal

roots and each root is2

b

a

.

(iii) If D < 0, the equation has no real roots.

3. (i) If 5p , then p –5

(ii) If – p –5, then p 5

(iii) If 2 4p , then either 2 or 2p p

(Important)

(iv) If 2 4p , then p lies between –2 and 2

i.e., –2 p 2 (Important)


Q.1. Find the nature of the roots of thefollowing quadratic equations. If the real rootsexist, find them :

(i)22 3 5 0x x

(ii) 23 4 3 4 0x x

(iii) 22 6 3 0x x

Sol. (i) Given equation is : 22 3 5 0x x− + =

This is of the form 2 0ax bx c+ + =

Where, 2, 3, 5= = − =a b c

Discriminant =2 4b ac

2( 3) 4 2 5

9 40 31 0

Hence, given quadratic equation has not real root.

(ii) Given equation is : 23 4 3 4 0x x− + =

This is of the form 2 0 ax bx c

Where, 3, 4 3, 4a b c

Discriminant = 2 4b ac

2( 4 3) 4 3 4

48 48 0

Thus, given equation has real and equal roots,which are given by

2 4

2

b b acx

a

− ± −=

( 4 3) 0 4 3 2

2 3 2 3 3

− − ±= = =

× ×Hence, roots of given quadratic equation are

2 2and

3 3.

(iii) Given equation is : 22 6 3 0x x− + =

This is of the form2 0ax bx c+ + =

Where, 2, 6, 3 a b c

Discriminant = 2 24 ( 6) 4 2 3b ac

36 24 12 0 Thus, given equation has real and distinct roots.

So,2 4

2

b b acx

a

( 6) 12 6 2 3

2 2 4

2[3 3] 3 3

4 2

So, roots are3 3 3– 3

and2 2

Hence, roots of given quadratic equation are

3 3 3 – 3and

2 2

.

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Q.2. Find the values of k for each of thefollowing quadratic equations, so that they havetwo equal roots.

(i)22 3 0x kx+ + =

(ii) ( 2) 6 0kx x − + =Sol. (i) Given equation is 22 3 0x kx+ + =

This is of the form 2 0ax bx c+ + =Where, 2, , 3a b k c= = =As, roots of the given quadratic equation are equal,

2 4 0b ac− =

2( ) 4 2 3 0k − × × = 2 24 0k − =

=2 24k = ± 24k

So, 2 6k = ±(ii) Given equation is : ( 2) 6 0kx x − + =

2 2 6 0kx kx− + =

This is of the form2 0ax bx c+ + =

Where, , 2 , 6a k b k c= = − =As, roots of the given quadratic equation are equal.

2 4 0b ac− = 2( 2 ) 4 6 0k k− − × × =

24 24 0k k− = 4 ( 6) 0k k − =Either, 4 0 0k k= ⇒ = or 6 0 6k k− = ⇒ = 0, 6.k =Q.3. Is it possible to design a rectangular

mango grove whose length is twice its breadth, and

the area is 800 m2 ? If so, find its length and

breadth. [V. Imp.]

Sol. Let breadth of rectangular grove = x m

Then, length of rectangular grove = 2x mArea of rectangular grove = length × breadth

= 2( 2 ) mx x×2 22 mx=

As per condition : 22 800x =

2 800

4002

x = =

400x = ± 20x = ±So, 20 [ 20]x x= ≠ − Breadth of rectangular grove = 20 m

Length of rectangular grove = 40 m

So, it is possible to design such a rectangle.

Q.4. Is the following situation possible? If so,

determine their present ages. The sum of the ages

of two friends is 20 years. Four years ago, the

product of their ages in years was 48. [V. Imp.]

Sol. Let age of first friend = x years. Then, age

of second friend = (20 – x) years.

Four years ago :

Age of first friend = ( 4) yearsx −

Age of second friend = (20 4)x− − years= (16 – x) years

Product of their ages = ( 4)(16 )x x− −

= 216 64 4 x x x2 20 64 x x

As per condition : 2 20 64 48x x− + − =

2 20 64 48 0x x− + − − =

2 20 112 0x x− + − =

2 20 112 0x x− + = ...(i)

This is of the form 2 0ax bx c+ + =Where, 1, 20, 112a b c= = − =

Discriminant 2 24 ( 20) 4 1 112b ac− = − − × ×400 448 48 0= − = − <

So, roots are not real. Then no real value of xsatisfies the quadratic equation (i).

Hence, given situation is not possible.

Q.5. Is it possible to design a rectangularpark of perimeter 80 m and area 400 m2? If so,find its length and breadth. [V. Imp.]

Sol. Let length of rectangular park = x m

Breadth of rectangular park = y m

Perimeter of rectangular park = 2(x + y) mArea of rectangular park = xy m2

As per condition : 2(x + y) = 80

80

402

x y+ = = y = 40 – x ...(i)

As per condition : xy = 400 ...(ii)Putting the value of y in (ii), we have

x(40 – x) = 400

240 400x x− = 240 400 0x x− − =

2 40 400 0x x− + =This is of the form 2 0ax bx c+ + =

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Where, 1, 40, 400a b c

Now, 2 24 ( 40) 4 1 400b ac− = − − × ×1600 1600 0= − =

So,2 4

2

b b acx

a

( 40) 0 4020

2 1 2

When, x = 20 then from (i), y = 40 – 20 = 20Length and breadth of the rectangular park are

each equal to 20 m. Given rectangular park can existif it is in the shape of a square.


Q.1. If the roots of 2 0 ax bx c are realand unequal, then :

(a) 2 4 0 b ac (b) 2 4 0 b ac

(c) 2 4 0 b ac (d) cannot say

Sol. (b) For real and unequal roots , 2 4 0b ac− >

Q.2. If2 0 ax bx c has equal roots, then :

(a) 2 4b ac > 0 (b)2 4b ac

(c)2 4a ab (d) none of these

Sol. (b) For equal roots discriminant = 0.

2 24 0 4 b ac b ac

Q.3. The discriminant of 2 2 0 bx acx bis :

(a) 22( )b ac− (b) 24( )b ac−(c) 24( )ac b− (d) 24 ( )ac b− −

Sol. (c) For 2 2 0bx ac x b− + =

Discriminant = 2( 2 ) 4. . .ac b b− −2 24 4 4( ) ac b ac b

Q.4. The discriminant of23 2 2 2 3 0x x+ − == 0 is :

(a) –16 (b) 32(c) –32 (d) none of these

Sol. (b) 2 24 (2 2) 4b ac− = + 3 2 3

8 24 32

Q.5. If 2 24 4 0a x abx k has equal roots,

then the value of k is :(a) a (b) a2

(c) b2 (d) b

Sol. (c) We have, 2 24 4 0a x abx k− + =For equal roots 2 4 0b ac− =

2 2( 4 ) 4.4 0 ab a k

2 2 216 16 0a b a k− = 2k b=Q.6. The quadratic equation

2 22( )p q x 2( ) 1 0 p q x has :

(a) equal roots(b) no real roots(c) real but not equal roots(d) none of these

Sol. (b) 2 2 24 [2( )] 4.2.( )b ac p q p q− = + − +

=2 2 24( ) 8( ) 4( ) 0 p q p q p q

Therefore, the given equation has no real roots.

Q.7. The quadratic equation whose roots arereal and equal is : [2011 (T-II)]

(a) 22 4 3 0x x (b)2 4 4 0x x

(c) 23 5 2 0x x (d) 2 2 2 – 6 0x x Sol. (b) Roots of quadratic equation will be real

and equal, if 2 4 0 b ac .

So, only in (b), 2 4 4 0x x

2 24 ( 4) 4.1.4 16 16 0b ac

Q.8. Which of the following equations has twodistinct real roots? [2011 (T-II)]

(a) 2 92 3 2 0

4x x

(b) x2 + x – 5 = 0

(c) x2 + 3x + 2 2 = 0

(d) 5x2 – 3x + 1 = 0Sol. (b) Roots of quadratic equation will be real

and distinct, if 2 4 0b ac .

So, only in (b), x2 + x – 5b2 – 4ac = (1)2 – 4.1.(–5) = 1 + 20 = 21 > 0

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Q.9. If the discriminant of 3x2 + 2x + a = 0 isdouble the discriminant of x2 – 4x + 2 = 0, then thevalue of a is : [2011 (T-II)]

(a) 2 (b) –2 (c) 1 (d) –1Sol. (d) (2)2 – 4.3.a = 2[(–4)2 – 4.1.2]

4 – 12a = 2(16 – 8)

4 – 12a = 16 –12a = 12 a = –1

Q.10. The quadratic equation with real andequal roots is :

(a) 22 5 7 0x x− + =

(b) 2 3 2 4 0x x− + =

(c) 22 2 2 2 0x x− + =(d) 2 3 9 0x x− + =

Sol. (c) Roots of quadratic equation will be real

and equal, if 2 4 0 b ac .

So, only in (c),22 2 2 2 0x x− + =

2 24 ( 2 2) 4. 2. 2 8 8 0b ac− = − − = − =

Q.11. If 2 2 5 4 0kx x has real and

equal roots, then the value of k is :

(a)4

5k = (b)

4

5k = −

(c)5

4k = (d)

5

4k = − .

Sol. (c) We have, 2 2 5 4 0kx x− + =

For real and equal roots, 2 4 0b ac− =2( 2 5) 4. .4 0 k

20 5

20 16 016 4

k k k

Q.12. The roots of 2 2 2 29 12 4 0p x pqrx q r− + =are :

(a) real and equal (b) real and distinct(c) not real (d) none of these

Sol. (a) We have, 2 2 2 29 12 4 0p x pqrx q r− + =

2 2 2 2 24 ( 12 ) 4.9 .4b ac pqr p q r− = − −

= 2 2 2 2 2 2144 144p q r p q r 0

So, roots are real and equal.

Q.13. If the quadratic equationabx2 + (b2 – ac)x – bc = 0 has equal roots then :

(a) 2b ac= (b) b c=

(c) 2b ac= − (d) ab ac=

Sol. (c) We have, 2 2( ) 0abx b ac x bc+ − − =For equal roots, discriminant = 0

2 2( ) 4 . 0b ac ab bc− + =

4 2 2 2 22 4 0b ab c a c ab c− + + = 2 2 2( ) 0 b ac b ac

Q.14. If 2 2 2 2 2( ) 0p x p q x q+ + + = has equal

roots, then 2 2p q− is equal to :

(a) 22q− (b) 22q

(c) 0 (d) none of these

Sol. (c) We have, 2 2 2 2 2( ) 0p x p q x q+ + + =Since it has equal roots, therefore, b2 – 4ac = 0

2 2 2 2 2( ) 4 0p q p q+ − =

4 4 2 2 2 22 4 0p q p q p q+ + − =

2 2 2 2 2( ) 0 0 p q p q

Q.15. If the quadratic equation

22 3 0x kx has equal roots, then the value of

k is :

(a) 5± (b) 6± (c) 2 6± (d) 3 2±

Sol. (c) Since, 22 3 0x kx− + = has equal roots

2 4 0 b ac2( ) 4 2 3 0 k

2 24 2 6k k= ⇒ = ± .

Q.16. If2 3

2 02

x k+ + = has real roots, then :

(a) 2 3k = (b) 2 3k = −(c) 2 3k = ± (d) 3k = ±

Sol. (c) If 2 32 0

2x k+ + = has equal roots, then

2 4 0 b ac

2 234 2 0 12

2 k k 2 3 k .

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Q.17. The roots of the equation

3x2 – 4x + 3 = 0 are : [2011 (T-II)](a) real and unequal (b) real and equal(c) imaginary (d) none of these

Sol. (c) b2 – 4ac = (–4)2 – 4. 3. 3 = 16 – 36 = –20 < 0Since b2 – 4ac < 0, therefore, roots are not real, i.e.,

imaginary.

Q.18. The value of k, for which the quadratic

equation 4x2 + 4 3 x + k = 0 has equal roots is :[2011 (T-II)]

(a) k = 2 (b) k = –2(c) k = –3 (d) k = 3

Sol. (d) For equal roots, b2 – 4ac = 0

(4 3 )2 – 4.4.k = 0

48 – 16 k = 0 k = 3

Q.19. The value of p for which the quadraticequation x(x – 4) + p = 0 has real roots, is :

[2011 (T-II)](a) p < 4 (b) p 4(c) p = 4 (d) none of these

Sol. (a) For real roots, b2 – 4ac > 0

(–4)2 – 4.1.p > 0 p < 4

Q.20. Write the nature of roots of quadratic

equation 24 4 3 3 0x x+ + = [2009]

Sol. We have, 24 4 3 3 0x x+ + =

D = 2 24 (4 3) 4.4.3 48 48 0b ac− = − = − =As D = 0, the equation has real and equal roots.

Q.21. For what value of k, are the roots of thequadratic equation [2008C]

23 2 27 0 x kx real and equal ?Sol. On comparing the given equation with

2 0ax bx c+ + = , we get

a = 3, b = 2k, c = 27Since, roots are real and equal. D = 0

2 24 0 (2 ) 4.3.27 0b ac k− = ⇒ − =

2 2 324

4 324 0 814

k k− = ⇒ = = 9k = ±

Q.22. For what value of k, are the roots of thequadratic equation

2 4 1 0kx x+ + = equal and real ? [2008C]Sol. On comparing the given equation with


a = k, b = 4, c = 1

Since, roots are real and equal.

D = 0

2 24 0 (4) 4( ) (1) 0b ac k− = ⇒ − =16 4k⇒ − = 0

16

4 16 44

k k−

− = − ⇒ = =−

Q.23. For what value of k, does2( 12) 2k x− + ( 12) 2 0k x− + = have equal roots?

[2008, 2011 (T-II)]Sol. On comparing the given equation with


a = k – 12, b = 2 (k – 12), c = 2Since, roots are real and equal, therefore,

D = 0 2 4 0 b ac2[2( 12)] k 4( 12)(2) 0k− − =

24( 12) 8( 12) 0k k− − − = 4( 12) ( 12 2) 0k k− − − =

( 12) ( 14) 0 12 or =14 k k k k .

Q.24. Determine the nature of the roots of thefollowing quadratic equations :

(i) 2 2 29 24−a b x abcdx 2 216 0,+ =c d

0,a 0b

(ii) 2 2 22( ) 2( ) 1 0 a b x a b x

(iii) 2( ) ( ) 0b c x a b c x a+ − + + + =Sol. (i) We have,

2 2 2 2 29 24 16a b x abcdx c d− + = 0,

0, 0 a b

Here,2 2A 9 ,= a b B 24 , abcd 2 2C 16 c d

2D = B 4AC

= 2 2 2 2 2{ 24( )} 4 9 16abcd a b c d− − × ×

= 2 2 2 2 2 2 2 2576 576a b c d a b c d− = 0

Since, D = 0, the roots of the equation are real andequal, i.e., the equation has a repeated root.

(ii) We have , 2 2 22( ) 2( ) 1 0a b x a b x+ + + + =

Here, 2 2A 2( ), B 2( ), C 1 a b a b

2D = B 4AC2 2 2[2( )] 4 2( ) 1a b a b= + − × + ×

= 2 2 2 24( 2 ) 8( )a b ab a b+ + − +

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= 2 2 2 24 4 8 8 8a b ab a b+ + − −

=2 24 4 8− − +a b ab

=2 2 24( 2 ) 4( )− + − = − −a b ab a b

Since D < 0, the equation has no real roots.

(iii) We have 2( ) ( ) 0b c x a b c x a+ − + + + =Here, A , B ( ), C b c a b c a

D =2B 4AC

=2{ ( )} 4( ) a b c b c a

=2 2 2 2 2 a b c ab bc 2 4 4 ca ab ac

= 2 2 2 2( )a b c ab bc ca+ + − − +

= 2 2 2 2 2 2 a b c ab bc ca

=2( ) 0a b c− + >

Since D > 0, the roots of the equation are real andunequal.

Q.25. Find the value of k for which the givenquadratic equations has real and distinct roots :

(i) 2 2 1 0kx x+ + =(ii)

24 3 9 0x kx− + =Sol. (i) We have, 2 2 1 0kx x+ + =Here, , 2, 1a k b c= = =

2D = 4 4 4 1 4 4b ac k k− = − × × = −For the equation having real and distinct roots,

D > 0 4 4 0 1 k k

(ii) We have, 24 3 9 0x kx− + =Here, 4, 3 , 9a b k c= = − =

2

2 2

D = 4

9 4 4 9 9 144

b ac

k kFor the equation having real and distinct roots,

2D > 0 9 144 0k⇒ − >

2 214416

9 k k Either k < – 4 or k > 4

Q.26. If – 4 is a root of the quadratic equationx2 + px – 4 = 0 and the quadratic equation x2 + px+ k = 0 has equal roots, find the value of k.

[2011 (T-II)]Sol. Since – 4 is a root of the equation x2 + px – 4 = 0

(– 4)2 + p(– 4) – 4 = 0 [A root alwayssatisfies the equation]

16 – 4p – 4 = 0 4p = 12 p = 3

The equation x2 + px + k = 0 has equal roots. Discriminant = 0 p2 – 4k = 0 9 – 4k = 0 [ p = 3]

k =9

4.

Q.27. Find the value of k for which the

quadratic equation 2(2 1) 2k x+ + ( 3)k x

( 5) 0k+ + = has real and equal roots. [Imp.]

Sol. The given equation is

2(2 1) 2( 3) ( 5) 0k x k x k+ + + + + =Here, 2 1, 2( 3), ( 5) a k b k c k

2 2D = 4 4( 3) 4 (2 1) ( 5)b ac k k k

= 2 24 36 24 8 40 4 20k k k k k+ + − − − −

=24 20 16k k− − + =

24( 5 4)k k− + −For the equation having equal roots,

2D = 0 5 4 0 k k

5 25 4 1 4

2k

− ± + × ×= 5 41

2k

− ±= .

Q.28. Find the values of k for which the

quadratic equation 2( 1) 2k x+ − (3 1)k x+

8 1 0k+ + = has real and equal roots. [Imp.]

Sol. We have, 2( 1) 2(3 1) 8 1 0k x k x k+ − + + + =

Here, 1, 2(3 1), 8 1 a k b k c k

2 2D 4 [ 2(3 1)] 4( 1)(8 1) b ac k k k

=2 24(3 1) 4(8 8 1) k k k k

=2 24(9 1 6 ) 4(8 9 1) k k k k

= 2 236 4 24 32 36 4k k k k+ + − − −

=24 12 4 ( 3)k k k k− = −

For the equation having equal roots,

D = 0 4 ( 3) 0k k⇒ − = 4 0, 3 0 k k

0, 3k k= = .

Q.29. In the following, determine the set ofvalues of k for which the given quadratic equationhas real roots :

(i)23 2 0 x x k (ii) 24 3 1 0 x kx

Sol. (i) We have, 23 2 0x x k+ + =Here, 3, 2,a b c k= = =

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D = 2 24 (2) 4 3 4 12b ac k k− = − × × = −The given equation will have real roots, if

D 0 4 – 12 0 4 12 k k

4 1

12 3k k≤ ⇒ ≤

(ii) We have, 24 3 1 0x kx− + =Here, 4, 3 , 1a b k c= = − =

2D = 4b ac− =

2( 3 ) 4 4 1k− − × × = 29 16k −The given equation will have real roots, if

2 2D 0 9 16 0 9 16k k≥ ⇒ − ≥ ⇒ ≥

2 16

9k ≥ ⇒

4 4,

3 3

−≤ ≥k k

Q.30. If the roots of the equation2( ) ( )b c x c a x− + − ( ) 0a b+ − = are equal,

then prove that 2b = a + c

Sol. We have, 2( ) ( ) ( ) 0b c x c a x a b− + − + − =Here, A , B , C b c c a a b

D = 2( ) 4( )( )c a b c a b− − − −For real and equal roots, we must have

2D = 0 ( – ) 4( )( ) 0 c a b c a b

2 2 22 4 4 4 4 0c a ca ab b ac bc+ − − + + − = 2 2 2(2 ) 2(2 ) 2(2 ) a b c ab bc 2 0 ca

2( 2 ) 0 2 0a b c a b c− + = ⇒ − + = 2b a c= + Proved.

Q.31. If the roots of the equation2 2 2 2 2( ) 2( ) ( ) 0a b x ac bd x c d aree

equal, prove thata c

b d . [Imp.]

Sol. We have,2 2 2 2 2( ) 2( ) ( ) 0a b x ac bd x c d+ − + + + =

Here, 2 2 2 2A , B 2( ),Ca b ac bd c d

2

2 2 2 2 2

D = B 4AC

= 4 ( 4( )( )ac +bd) a b c d

For real and equal roots, we must have D = 0.Now, D = 0

2 2 2 2 24( ) 4 ( )( ) 0 ac+bd a b c d

2 2 2 2 2 22a c b d abcd a c+ + −2 2 2 2 2 2 0a d b c b d− − − =

2( ) 0ad bc ad bc− = ⇒ =

a c

b d= Proved.

Q.32. If p, q, are real and p q , then showthat the roots of the equation

2( ) 5( )p q x p q 2( ) 0x p q are realand unequal. [Imp.]

Sol. We have 2( ) 5( ) 2( ) 0p q x p q x p q− + + − − =Here, , 5( )= − = +a p q b p q and c = –2 (p – q)

2 2D = 4 25( )b ac p q− = +4( ) { 2( )}p q p q− − × − −

2 2= 25 ( ) 8( ) p q p q

Clearly, 225( ) 0p q+ > and 28( ) 0p q− >[ p q≠ ]

2 2D = 25 ( + ) 8( ) p q p q > 0

Hence, roots of the given equation are real andunequal. Proved.

Q.33. Show that the equation2 22( ) 2( ) 1 0 a b x a b x has no

real roots, when a b≠ .Sol. The given equation is

2 2 22( ) 2( ) 1 0a b x a b x+ + + + =

Here 2 2A 2( ), a b B = 2 (a + b) and C = 1

2 2 2 2D = B 4AC 4( + ) –8 ( ) a b a b

= 2 2 24( 2 ) 4( ) 0− + − = − − <a b ab a b

When a – b 0So, the given equation has no real roots, when

a b≠ . Proved.

Q.34. Prove that both the roots of the equation

( ) ( ) ( )x a x b x b− − + − ( )x c + (x – c)(x – a) = 0are real but they are equal only when a = b = c.

Sol. The given equation may be written as2 2( ) ( ) x a b x ab x b c x bc

2 ( ) 0 x c a x ca

23 2 ( ) ( ) 0x x a b c ab bc ca− + + + + + =

D = 24( ) 12( )a b c ab bc ca+ + − + +

=2 24( )a b c ab bc ca2 + + − − −

=2 2 22(2 2 2 2 2 2 )a b c ab bc ca+ + − − −

2 2 22[( ) ( ) ( ) ] 0a b b c c a= − + − + − ≥2 2 2[ ( ) 0, ( ) 0 and ( ) 0] a b b c c – a

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Choose the correct option (Q. 1 – 9)

1. Discriminant of 2 2 0x px q is :

(a) 8p q (b)2 8p q

(c)2 8p q (d) 2 8q p

2. If the equation2 4 0 x x k has real and

distinct roots, then :(a) k < 4 (b) k > 4 (c) 4k (d) 4k

3. If the equation 22 5 3 6 0x x has equalroots then the roots are :

(a)3

2 3,2

(b)3

, 2 32

(c) 3, 2 (d) 3, 2

4. The quadratic equation ax2 + bx + c = 0,a 0 has two distinct real roots, if :

(a) 2 4 0 b ac (b)2 4 0 b ac

(c) 2 4 0 b ac (d) none of these

5. If2 0ax bx c has equal roots, then c is

equal to :

(a)2

b

a(b)

2

b

a(c)

2

4

b

a(d)

2

4

b

a

6. If the equation 29 6 4 0x kx has equalroots, then the roots are : [2011 (T-II)]

(a)2

3 (b)

3

2 (c) 0 (d) 3

7. The roots of the equation 22 5 5 0 x xare :

(a) real and distinct (b) not real(c) real and equal (d) real and unequal

This shows that both the roots of the given equationare real. Proved.

For equal roots, we must have D = 0.Now, D 0

2 2 2( ) ( ) ( ) 0 a b b c c a

( ) 0, ( ) 0 a b b c and ( – ) 0c a a = b = c

Hence, the roots are equal only when a = b = cProved.

Q.35. Determine the positive value of k for

which both the equations 2 64 0x kx and2 8 0x x k will have real roots.

Sol. We have, 2 64 0 x kx ...(i)

and, 2 8 0x x k− + = ...(ii)

Let D1 and D2 be the discriminants of equations(i) and (ii) respectively. Then

21D 4 64k= − × 2 256k= − and

22D ( 8) 4 64 4k k= − − = −

Both the equations will have real roots, if

1 2D 0 and D 0≥ ≥

2 256 0 and 64 – 4 0 k k

2 256 and 64 4k k≥ ≥ 16 and 16 k k k = 16

Hence, both the equations will have real roots whenk = 16.

Q.36. If the roots of the equation

2 2 2 2( ) 2( ) ( ) 0 c ab x a bc x b ac are real

and equal, then show that either a = 0 ora3 + b3 + c3 = 3abc.Sol. Given equation is

2 2 2 2( ) 2( ) ( ) 0c ab x a bc x b ac− − − + − =

Here, 2 2A , B 2( ), c ab a bc 2C b ac D = B2 – 4AC

= 2 2 2 24( ) 4 ( ) ( )a bc c ab b ac− − × − × −

= 4 2 2 24( 2 )a b c a bc+ −2 2 3 3 24( )c b ac ab a bc− − − +

= 4 2 2 24( 2a b c a bc+ −2 2 3 3 2 ) c b ac ab a bc

= 3 3 34 ( 3 )a a b c abc+ + −For real and equal roots, D = 0

3 3 34 ( 3 ) 0a a b c abc+ + − =

3 3 30 or 3 0a a b c abc= + + − =

3 3 30 or 3a a b c abc= + + = Proved.

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8. If the roots of the equation

2 2 2( ) 2a b x b+ − 2 2( ) ( ) 0a c x b c+ + + = are

equal, then :

(a) 2 b a c (b) 2 b ac

(c)2

acb

a c(d) b ac

9. If 1x is a common root of the equations

2 3 0ax ax and 2 0x x b , then ab isequal to :

(a) 3 (b) 3.5 (c) 6 (d) –3

10. Write the discriminant to each of thefollowing quadratic equations :

(i) 2 2 1 0 x x (ii)2 4 1 0 x x

(iii) 23 2 1 0 x x (iv) 2 26 6 x ax a

(v) 22 3 2 0 x x

(vi) 2 4 0 x x b

(vii) 2(2 ) 2 a b x x ab

(viii) ( 1)(2 1) 0 x x

11. Determine the nature of the roots of the

following quadratic equations (Don't solve) :

(i) 23 4 5 0 x x (ii)24 2 3 0 x x

(iii) 2 30 225 0 x x

(iv) 22 3 2 0 x x

(v) 22 3 5 0 x x

(vi) 24 20 25 0 x x

(vii) 22 7 8 0 x x

(viii) 25 2 0 x x

(ix) 2 6 9 0 x x (x) 23 2 5 0 x x

12. Find the values of k for which the roots arereal and equal in each of the following equations :

(i)22 40 25 0 kx x

(ii) 2 2(5 2 ) 3(7 10 ) 0 x k x k

(iii) 2(3 1) 2( 1) 0 k x k x k

(iv) 2 21 4 kx kx x x

(v) 2( 1) 2( 3) ( 8) 0 k x k x k

(vi)2 2 7 12 0 x kx k [2011 (T-II)]

(vii) 2 25 4 2 (4 2 1) 0 x x k x x

(viii) 2(4 ) (2 4) (8 1) 0 k x k x k

13. Show that the equation 2 4 0x ax+ − = hasreal and distinct roots for all real value of a.

14. If the roots of the equations2 2 0+ + =ax bx c and 2 2 0 bx ac x b are

simultaneously real, then prove that 2 b ac .[Imp.]

15. If a, b, c are real numbers such that 0≠acthen show that at least one of the equations

2 0 ax bx c and 2 0 ax bx c has real

roots.

16. If the equations

(1 + m2)x2 + 2mcx + (x2 – a2) = 0, has equalroots, prove that c2 = a2(1 + m2). [Imp.]

17. If p, q, r and s are real numbers such that2( ) pr q s , then show that at least one of the

equations 2 0 x px q and 2 0 x rx s has

real roots. [Imp.]

18. Prove that the equation2 2 2( ) 2 x a b x 2 2( ) ( ) 0 ac bd c d

has no real root, if ad bc . [Imp.]

19. If the roots of the equationx2 + 2cx + ab = 0 are real and unequal, prove that

the equation 2 2 2 22( ) 2 0 x a b x a b chas no real roots. [Imp.]


Q.1. There are three consecutive positive

integers such that the sum of square of the first

and the product of the other two is 154. What are

the integers? [2011 (T-II)]

Sol. Let the three consecutive positive integers be

, 1, 2 x x x .

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According to the question,2 ( 1) x x ( 2) 154 x

2 2 3 2 154 0 x x x

22 3 152 0 x x

22 19 16 152 0 x x x

(2 19) 8(2 19) 0 x x x

(2 19)( 8) 0 x x

19

or 82

x x

Rejecting19

2

x , because x is an integer

x = 8Hence, the three consecutive positive integers are

8, 1 8 1 9 x x , and 2 8 2 10 x i.e., 8,

9 and 10.

Q.2. The difference of the squares of two

numbers is 45. The square of the smaller number

is 4 times the larger number. Find the numbers.

[2007]Sol. Let the larger number be x and the smaller

number be y. Then, 2 2 45− =x y ...(i)

and 2 4=y x ...(ii)

Substituting 2 4y x= in (i), we have 2 4 45− =x x

2 4 45 0− − =x x

2 9 5 45 0− + − =x x x

( 9) 5 ( 9) 0− + − =x x x

( 9)( 5) 0− + =x x

9 0 or +5= 0− =x x

9 or 5= = −x x

Smaller number = 36 or 20− . [From (ii)]

Smaller number = 6 [ 20 is not real]−

Larger number = 9 and smaller number = 6

Q.3. A two digit number is such that the productof its digits is 14. If 45 is added to the number, thedigits interchange their places. Find the number.

Sol. Let the number be 10x + y.

Product of the digits = xy

xy = 14 (Given)

14

yx

= ...(i)

According to the question,10 45 10 x y y x

9 9 45 0x y− + = 5 0x y− + =

14

5 0xx

− + = [From (i)]

2 14 5 0x x− + =

2 5 14 0x x+ − =

( 7)( 2) 0x x+ − =

2 and 7x x= = − [x = –7 is rejected since a

digit of a number cannot be negative]

x = 2 14 14

72

yx

= = =

The number is (10x + y) = 10(2) + 7 = 27

Q.4. The sum of two numbers is 16. The sum

of their reciprocals is1

3. Find the numbers.

[2005]Sol. Let the required numbers be x and 16 – x. Then

1 1 1

16 3x x+ =

−

16 1

(16 ) 3

x x

x x

− +=

−

16 1

(16 ) 3x x=

−

248 16x x= −

2 16 48 0x x− + = 2 12 4 48 0x x x− − + = ( 12) 4( 12) 0x x x− − − = ( 12)( 4) 0x x− − = 12 or 4x x= =Hence, the two numbers are 12 and 4.

Q.5. The denominator of a fraction is one morethan twice the numerator. If the sum of the fraction

and its reciprocal is 216

21, find the fraction.

[2011 (T-II)]Sol. Let the numerator of the fraction be x. Then,

Denominator = 2x + 1

Fraction =2 1

x

x

Reciprocal of the fraction =2 1x

xIt is given that the sum of the fraction and its

reciprocal is 216

21.

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2 1 162

2 1 21

x x

x x

2 2(2 1) 58

(2 1) 21

x x

x x

2

25 4 1 58

212

x x

x x

21(5x2 + 4x + 1) = 58(2x2 + x)

105x2 + 84x + 21 = 116x2 + 58x

11x2 – 26x – 21 = 0

11x2 – 33x + 7x – 21 = 0

11x(x – 3) + 7(x – 3) = 0

(11x + 7)(x – 3) = 0

x = 3 or x =7

11 x = 3 [ x is a natural number x > 0]

Hence, fraction =

3

2 1 7

x

x.

Q.6. Find the whole number which whendecreased by 20 is equal to 69 times the reciprocalof the number.

Sol. Let the required number be x .

Its reciprocal =1

x.

Using the given information, we get

1( 20) 69x

x

− =

( 20) 69 x x 2 20 69 x x

2 20 69 0x x− − = 2 23 3 69 0x x x− + − = ( 23) 3( 23) 0x x x− + − = ( 23) ( 3) 0x x− + = 23 0 or + 3 = 0 x x

⇒ 23 or 3x x= = −But 3x ≠ − , since x is a whole number..

Hence, the required number is 23.

Q.7. Out of a number of Saras birds, one fourth

the number are moving about in lotus plants,1

9th

coupled (along) with1

4as well as 7 times the

square root of the number move on a hill , 56 birds

remain in vakula trees. What is the total number ofbirds ? (Mahavira around 850 A.D.)

Sol. Let the total number of birds be 2x .

Birds moving about in lotus plants =2

4

x

Birds moving on a hill =2 2

79 4

x xx+ +

Birds remain in vakula trees = 56

2 2 2

27 564 9 4

x x xx x+ + + + =

2 2

27 562 9

x xx x+ + + =

2 2 29 2 126 1008 18x x x x+ + + =

27 126 1008 0x x− − =

2 18 144 0 x x

2 24 6 144 0x x x− + − = ( 24) 6( 24) 0x x x− + − = ( 24) ( 6) 0x x− + = (x – 24) = 0 or ( 6) 0x

24 or 6x x= = − Total number of birds = (24)2 = 576or (–6)2 = 36

We reject 2 36x = , since it does not satisfy the

given conditions.Hence, total number of birds = 576.

Q.8. The sum of two numbers is 8. Determine

the numbers if the sum of their reciprocals is8

15.

[2009, 2011 (T-II)]Sol. Let one of the numbers be x.Then, another number = 8 – x

Also,2

1 1 8 8 8

8 15 158

x x

x x x x

− ++ = ⇒ =

− −

2 28 15 8 15 0x x x x− = ⇒ − + =

2 5 3 15 0x x x− − + = ( 5) 3( 5) 0x x x⇒ − − − = ( 5) ( 3) 0 5 or 3x x x x− − = ⇒ = = .Hence, the numbers are 5 and 3.

Q.9. The numerator of a fraction is one lessthan its denominator. If three is added to each of

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the numerator and denominator, the fraction is

increased by3

28. Find the fraction. [2007]

Sol. Let denominator be x, then numerator is (x – 1).

Fraction =1x

x

−

According to question,1 3 1 3

3 28

x x

x x

− + −= +

+

2 28 28 3

3 28

x x x

x x

28 ( 2) ( 3) (31 28)x x x x+ = + −

2 228 56 31 28 93 84 x x x x x

2 228 56 31 28 93 84 0x x x x x+ − + − + =

– 23 9 84 0x x− + = 2 3 28 0x x+ − =

2 7 4 28 0x x x+ − − = ( 7) 4( 7) 0 ( 7)( 4) 0 x x x x x

7 0 or 4 0x x+ = − = 7 or 4x x= − =

When 4x , fraction =4 1 3

4 4

−= .

When 7x = − , fraction =7 1 8 8

7 7 7

− − −= =

− −

[Rejecting8

7, since,numerator is one less than its

denominator]

Fraction =3

4.

Q.10. The difference of two numbers is 5 and

the difference of their reciprocals is1

10. Find the

numbers. [2007]Sol. Let the larger number be x and the smaller

number be y.

According to question, 5x y− = .....(i)

1 1 1

10y x− =

⇒ 1

10

x y

xy

− = [ If x > y then1

x<

1

y]

5 1 50

5010

xy yxy x

...(ii)

Putting the value of y in (i), we get

250 50 55

1

xx

x x

−− = ⇒ =

2 250 5 5 50 0x x x x− = ⇒ − − =

2 10 5 50 0x x x− + − = ( 10) 5( 10) 0x x x− + − =

( 10)( 5) 0 10 or 5x x x x− + = ⇒ = = −Putting these values of x in (ii)

50 5010 or 5

5 10

y y y y

Numbers are : – 5, – 10 or 10, 5.

Q.11. By increasing the list price of a book byRs 10 a person can buy 10 books less for Rs 1200.Find the original list price of the book. [2007]

Sol. Let original price of the book = Rs x andincreased price of the book = Rs (x +10).

Total amount = Rs 1200

According to question,1200 1200

1010x x

− =+

1200 12000 1200 10

( 10) 1

x x

x x

+ −=

+

12000

10 10 ( 10) 12000( 10)

x xx x

= ⇒ + =+

2( 10) 1200 10 1200 0x x x x+ = ⇒ + − =

2 40 30 1200 0x x x+ − − =( 40) 30( 40) 0x x x⇒ + − + =

( 40) ( 30) 0x x+ − =40 or = 30⇒ = −x x

[Rejecting –ve value, because price cannot be –ve]

The original price of the book = Rs 30.

Q.12. A 2-digit number is such that the productof its digits is 18. When 63 is subtracted from thenumber, the digits interchange their places. Findthe number. [2006, 2011 (T-II)]

Sol. Let unit’s digit be x and ten's digit be18

x.

Original number18

10

xx

180 x

x...(i)

and interchanged number =18

10xx

+ .

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According to the question,

18063x

x+ −

1810x

x= +

18 180

10 63 0x xx x

+ − − + =

2 210 18 180 63

0x x x

x

+ − − +=

2

29 63 1620 9 63 162 0

x xx x

x

+ −= ⇒ + − =

2 7 18 0x x+ − =

2 9 2 18 0x x x+ − − = ( 9) 2( 9) 0x x x⇒ + − + = ( 9)( 2) 0 9 0 or 2 0x x x x+ − = ⇒ + = − = 9 or 2 x x

But digits cannot be –ve.

x = 2.Putting the value of x in (i), we get

Original number =180

2 2 90 922

+ = + = .

Q.13. A train covers a distance of 90 km at auniform speed. Had the speed been 15 km/hr more,it would have taken 30 minutes less for the journey.Find the original speed of the train. [2006]

Sol. Let the original speed of the train be x km/hr.Then the increased speed of the train = (x + 15) km/hr.

Distance = 90 kmAccording to the question,

90 90 1

15 2x x− =

+1

30 mins. hr.2

90( 15) 90 1

( 15) 2

+ −=

+x x

x x

90 90 15 90 1

( 15) 2

x x

x x

+ × −=

+

2( 15) 90 15 2 15 2700x x x x+ = × × ⇒ + =

2 15 2700 0 x x

2 60 45 2700 0 x x x

( 60) 45( 60) 0x x x+ − + =( 60) ( 45) 0x x⇒ + − =

45 0 or 60 0x x− = + = 45 or 60x x= = −But speed of the train cannot be –ve.

The original speed of the train = 45 km/hr..

Q.14. Seven years ago Varun's age was fivetimes the square of Swati's age. Three years henceSwati's age will be two-fifth of Varun's age. Findtheir present ages. [2006]

Sol. Let the present age of Varun = x years and the

present age of Swati = y yearsFirst condition : ( 7 years ago)x – 7 = 5(y – 7)2 ... (i)Second condition : (3 years hence)

2( 3) ( 3)

5x y+ = +

2 6 5 15 2 5 9x y x y+ = + ⇒ = +

5 9

2

yx ... (ii)

Putting the value of x in (i) , we get

25 97 5( 7)

2

yy

25 9 145( 14 49)

2

yy y

25( 1)

5( 14 49)2

yy y

−= − +

21 14 49

2 1

y y y

22 28 98 1y y y− + = −22 28 98 1 0 y y y

2y2 – 29y + 99 = 0 2y2 – 18y – 11y + 99 = 0

2 ( 9) 11( 9) 0y y y− − − =

( 9) (2 11) 0 y y

9 0 or 2 11 0y y− = − =

11

9 or2

y y= =

[Rejecting y =11

2as it is fractional]

Swati's present age (y) = 9 years.Putting this value of y in (ii), we get

5 9 9 5427

2 2x

× += = =

Varun's present age (x) = 27 years.

Q.15. The sum of ages of father and his son is45 years. 5 years ago, the product of their ageswas 124. Determine their present ages.

[2011 (T-II)]

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Sol. Let the present age of father be x years. Then,

Son's present age = (45 – x) yearsFive years ago :Father's age = (x – 5) yearsSon's age = (45 – x – 5) years = (40 – x) yearsIt is given that five years ago, the product of their

ages was 124. (x – 5)(40 – x) = 124 40x – x2 – 200 + 5x = 124 x2 – 45x + 324 = 0 x2 – 36x – 9x + 324 = 0 x(x – 36) – 9(x – 36) = 0 (x – 36)(x – 9) = 0 x = 36 or x = 9When x = 36, we haveFather's present age 36 yearsSon's present age = 9 yearsWhen x = 9, we haveFather's present age = 9 yearsSon's present age = 36 yearsClearly, this is not possibleHence, Fathers present age = 36 years and Son's

present age = 9 years.

Q.16. A peacock is sitting on the top of a pillar,which is 9 m high. From a point 27 m away fromthe bottom of the pillar, a snake is coming to itshole at the base of the pillar. Seeing the snake thepeacock pounces on it. If their speeds are equal, atwhat distance from the hole is the snake caught?

[2007]Sol. Let PH be the pillar and at the point P a peacock

is sitting on the top of pillar.Let the distance from the hole to the place where

snake is caught = x mP be the top of the pillar and S be the point from

where the snake starts moving.

SC = (27 – x) mSC = PC = (27 – x) m [ Their speeds are equal]

In right PHC , PH2 + CH2 = PC2

[Pythagoras theorem]

S CH

P

(27

–)m

x

9 m

(27 – ) mx

27 m

x m

2 2 2(9) ( ) (27 )x x+ = −2 281 729 54 x x x

648

54 729 81 648 12 m54

x x= − = ⇒ = =

Hence, the required distance, x = 12 m.

Q.17. The difference of two numbers is 4. If

the difference of their reciprocals is4

21, find the

two numbers. [2008]

Sol. Let the larger number be x and the smaller

number be y.

According to the question, 4x y− = ...(i)

1 1 4

21y x− = [ If x > y then

1 1

x y< ]

4

21

x y

yx

−=

4 4

21xy= [From (i)]

xy = 21 ⇒21

yx

= ...(ii)

Putting the value of y in (i), we get

221 214 4

xx

x x

−− = ⇒ =

2 21 4x x− = ⇒ 2 4 21 0x x− − =

2 7 3 21 0x x x− + − =( 7) 3( 7) 0x x x⇒ − + − =

( 3)( 7) 0x x+ − =3 0 or 7 0 x x

3x = − or 7x =Putting the value of x in (ii) , we get

21 217 or = 3

3 7y y= = − =

− Numbers are : –3, – 7 or 7, 3.

Q.18. In a class test, the sum of the marks

obtained by P in Mathematics and Science is 28.

Had he got 3 more marks in Mathematics and 4

marks less in Science, the product of marks

obtained in the two subjects would have been 180.

Find the marks obtained in the two subjects

separately. [2008]Sol. Let marks obtained by P in Mathematics = x

and marks obtained by P in Science = (28 – x)According to the question,

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( 3)[(28 ) 4] 180 x x

( 3)(24 ) 180x x+ − =

224 72 3 180x x x⇒ − + − =

2 21 72 180 0x x− + + − =2 21 108 0x x⇒ − + − = 2 21 108 0x x− + =

2 12 9 108 0x x x− − + =

( 12) 9( 12) 0x x x− − − =( 9) ( 12) 0x x⇒ − − =

9 0 or 12 0x x− = − = 9 or 12x x= =

When x = 9, When x = 12,Mathematics marks Mathematics marks

= x = 9 = x = 12and Science marks and Science marks

= 28 – x = 28 – x= 28 – 9 = 28 –12= 19 = 16

Q.19. The sum of the areas of two squares is640 m2. If the difference in their perimeters be64 m, find the sides of the two squares. [2008]

Sol. Let the side of the larger square = x cm and the

side of the smaller square = y cmAccording to the question,

4 4 64x y− = [ Perimeter of sq. = 4 × side]

16x y− = 16x y= + ... (i)

Also, 2 2 640x y+ =[Area of the square = (side)2]

2 2(16 ) 640y y+ + = [From (i)]

2 2256 32 640 0y y y+ + + − =

22 32 384 0y y+ − =

2 16 192 0y y+ − =

2 24 8 192 0y y y+ − − =( 24) 8( 24) 0y y y⇒ + − + =

( 24) ( 8) 0 24 or 8 y y y y

Since, side of a square can't be –ve.

y = 8 cmNow, putting the value of y in (i), we getx = (16 + 8) cm = 24 cm

Side of the larger square = 24 cmand side of the smaller square = 8 cm.

Q.20. A motor boat whose speed is 18 km/h in

still water takes one hour more to go 24 km

upstream than to return downstream to the same

spot. Find the speed of the stream.

[2008, 2011 (T-II)]

Sol. Let the speed of the stream = x km/hr

Then the speed of boat upstream

= (18 – x) km/hr and the speed of boat downstream

= (18 + x) km/hrDistance = 24 km

According to the question,24 24

118 18

x x

DistanceTime =

Speed

24[(18 ) (18 )]

1(18 )(18 )

x x

x x

+ − −=

− +

2 224 2 (18) x x 248 324 x xx2 + 48x – 324 = 0

x2 + 54x – 6x – 324 = 0

( 54) 6( 54) 0x x x+ − + =

( 54) ( 6) 0x x⇒ + − = 54 0 or 6 0x x 54 or 6x x

Speed of the stream = 6 km/hr[ Speed of boat cannot be –ve]

Q.21. The sum of the squares of twoconsecutive odd numbers is 394. Find the numbers.

[2009]Sol. Let the consecutive odd numbers be x and

(x + 2).

Then, 2 2( 2) 394x x+ + =

2 2 4 4 394 x x x

22 4 390 0x x+ − =

2 2 195 0x x+ − =

2 13 15 195 0x x x− + − = ( 13) 15( 13) 0x x x− + − =

( 13) ( 15) 0x x− + =

13 or 15 x x

For x = 13, (x + 2) = 15,and for x = – 15, (x + 2) = –13Hence, the required odd numbers are 13 and 15 or

–15 and –13.

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Q.22. Places A and B are 100 km apart on ahighway. One car starts from A and another fromB at the same time. If the cars travel in the samedirection at different speeds, they meet in 5 hours.If they travel towards each other, they meet in 1hour. What are the speeds of the two cars?[2009]

Sol. Let the speed of first car, starting fromA = x km/hr

And the speed of the second car, starting fromB = y km/hr.

When moving in the same direction, let they meetat C.

100 km

B CA

Distance travelled by first car in 5 hours = AC = 5xDistance travelled by second car in 5 hours

= BC = 5yBut, AC = AB + BC

5 100 5x y= + 20x y= + ...(i)

When moving in the opposite directions, let theymeet at D.

100 km

BDA

Distance travelled by first car in 1 hour = AD = xDistance travelled by second car in 1 hour = BD = yAD + BD = AB x + y = 100 ...(ii)Substituting x = 20 + y from (i) in (ii), we have

(20 ) 100y y+ + = ⇒ 20 + 2y = 100

2y = 100 – 20 = 80 ⇒ y = 40 km/hrOn putting y = 40 in (i), we getx = 20 + 40 = 60 km/hr.Hence, the speed of first car = 60 km/hr and the

speed of the second car = 40 km /hr.

Q.23. A trader bought a number of articles forRs 900. Five articles were found damaged. He soldeach of the remaining articles at Rs 2 more thanwhat he paid for it. He got a profit of Rs 80 on thewhole transaction. Find the number of articles hebought. [2009]

Sol. Let no. of articles bought= x and cost of each article = y.

According to the question,

1st condition : 900900 xy y

x

2nd condition :900

( 5) 2 900 80xx

− + = +

900 2

( 5) 980x

xx

+ − =

2900 2 4500 10

980x x x

x

+ − −=

22 890 4500 980x x x+ − =

22 90 4500 0x x− − =2 45 2250 0x x⇒ − − =

2 75 30 2250 0x x x− + − =( 75) 30( 75) 0x x x⇒ − + − =

( 75) ( 30) 0x x 75 or 30x x⇒ = = −[x = –30 is rejected as it is –ve]

x = 75

No. of articles = 75

Q.24. Two years ago, a man's age was threetimes the square of his son's age. In three yearstime, his age will be four times his son's age. Findtheir present ages. [2009]

Sol. Let present age of son = x yearsTwo years ago,Age of son = (x – 2) yearsAge of man = 3 (x – 2 )2 years

Man's present age = 23( 2) 2 x

2 23( 4 4) 2 (3 12 14) x x x x years

After 3 years,

The age of son = ( 3)yearsx +

Age of man =2(3 12 14) 3x x− + +

= (23 12 17x x− + ) years

According to given condition,23 12 17 4( 3)x x x− + = +

23 12 17 4 12x x x− + = + 23 16 5 0 ( 5) (3 1) 0x x x x− + = ⇒ − − =

1

5 or3

x x= =

Since,1

3x = is not possible, therefore, 5x =

The present age of man

= 23 12 14x x− +23 5 12 5 14 = 29 years

And the present age of son = x = 5 years.

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Q.25. The age of a father is twice the squareof the age of his son. Eight years hence the age ofthe father will be 4 years more than three times theage of his son. Find their present ages. [2009]

Sol. Let the present age of son be x years.

Then, age of father = 2x2

After 8 years, age of son = (x + 8) years

And age of father =2(2 8) yearsx

According to given condition,

22 8 3( 8) 4 x x

22 8 3 24 4x x+ = + +

22 3 20 0x x− − = 22 8 5 20 0x x x− + − = 2 ( 4) 5( 4) 0x x x− + − =

( )2 5 ( 4) 0x x+ − = 2 5 0 or 4 0+ = − =x x

5or 4

2x x⇒ = − =

5

2x

−= [Rejected]

Hence, present age of son = 4 years.And, present age of father = 2(4)2 = 32 years.

Q.26. A girl is twice as old as her sister. Fouryears hence, the product of their ages (in years)will be 160. Find their present ages. [2010]

Sol. If age of sister is x, then age of the girl willbe 2x. 4 years hence, age of girl = 2x + 4 and age ofsister = x + 4

According to the question, (2 4) ( 4) 160x x+ + =

22 4 8 16 160x x x⇒ + + + =

2 22 12 144 0 6 72 0x x x x+ − = ⇒ + − =

2 12 6 72 0x x x+ − − =

( 12) 6( 12) 0x x x⇒ + − + =

( 12)( 6) 0 12 or = 6 x x x x

We reject x = – 12 as age cannot be –ve.

Present age of sister = x = 6 years

And present age of girl = 2x = 2 × 6 = 12 years

Q.27. A train travels 300 km at a uniformspeed. If the speed of the train had been 5 km/hourmore, it would have taken 2 hours less for the samejourney. Find the usual speed of the train.

[2011 (T-II)]

Sol. Let the usual speed of the train be x km/hr .

Then, time taken to cover 300 km =300

xkm.

Time taken to cover 300 km when the speed is

increased by 5 km/hr =300

5x km.

It is given that the time to cover 300 km is reducedby 2 hrs.

300 300

5x x

= 2

300 5 300

5

x x

x x

= 2

2

300 1500 300

5

x x

x x

= 2

2x2 + 10x = 1500 x2 + 5x – 750 = 0 x2 + 30x – 25x – 750 = 0 x(x + 30) – 25(x + 30) = 0 (x + 30)(x – 25) = 0 x = –30 or x = 25But, x cannot be negative. Therefore, x = 25.Hence, the usual speed of the train is 25 km/hr.

Q.28. A plane left 30 minutes later than theschedule time and in order to reach its destination1500 km away in time, it has to increase its speedby 250 km/hr from its usual speed. Find its usualspeed. [2011 (T-II)]

Sol. Let the usual speed of the plane be x km/hr.Then, time taken to cover 1500 km with the usual

speed =1500

xhrs.

Time taken to cover 1500 km with the speed of

(x + 250) km/hr =1500

250x

1500 1500 1

250 2x x

1500 1500 1

250 2x x

1500 1500 250 1500 1

( 250) 2

x x

x x

2

1500 250 1

2250x x

750000 = x2 + 250x

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x2 + 250x – 750000 = 0 x2 + 1000x – 750x – 750000 = 0 x(x + 1000) – 750(x + 1000) = 0 (x + 1000)(x – 750) = 0 x = –1000 or x = 750 x = 750 [ speed cannot be negative]Hence, the usual speed of the plane is 750 km/hr.

Q.29. Two pipes can together fill a tank in 31

13minutes. If one pipe takes 3 minutes more than theother to fill it, find the time in which each pipe canfill the tank. [2011 (T-II)]

Sol. Suppose the faster pipe takes x minutes to fillthe cistern. Therefore, the slower pipe will take (x + 3)minutes to fill the cistern.

Since the faster pipe takes x minutes to fill thecistern.

Portion of the cistern filled by the faster pipe in

one minute =1

x Portion of the cistern filled by the faster pipe in

40

13minutes =

1 40 40

13 13x x

Similarly, portion of the cistern filled by the slower

pipe in40

13minutes =

1 40 40

13 13 13( 3)x x

It is given that the cistern is filled in40

13minutes.

40 401

13 13 3x x

1 1 13

3 40x x

3 13

( 3) 40

x x

x x

40(2x + 3) = 13x(x + 3)

80x + 120 = 13x2 + 39x

13x2 – 41x – 120 = 0

13x2 – 65x + 24x – 120 = 0

13x(x – 5) + 24(x – 5) = 0

(x – 5)(13x + 24) = 0

x – 5 = 0 or 13x + 24 = 0

x = 5 or x =24

13

x = 5 [ x > 0]

Hence, the faster pipe fills the cistern in 5 minutesand the slower pipe takes (5 + 3) = 8 minutes to fill thecistern.

Q.30. Some students planned a picnic. Thebudget for food was Rs 480. But 8 of them failed togo, the cost of food for each member increased byRs 10. How many students attended the picnic?

[2011 (T-II)]Sol. Suppose x students attended the picnic.

Then, expense on food per head = Rs480

xNumber of students who planned the picnic = (x + 8)Then, in this case, expense on food per head

= Rs480

8x But, the difference in these expenses = Rs 10

480 480

108x x

1 1 10

8 480x x

8 1

( 8) 48

x x

x x

48 × 8 = x2 + 8x x2 + 8x – 384 = 0 x2 + 24x – 16x – 384 = 0 x(x + 24) – 16(x + 24) = 0 (x + 24)(x – 16) = 0 x = –24 or x = 16Rejecting x = –24, because number of students

cannot be negative.Hence, 16 students attended the picnic.

Q.31. The product of Tanay's age (in years)five years ago and his age ten years later is 16.Determine Tanay's present age. [2011 (T-II)]

Sol. Let Tanay's present age be x years.

His age 5 years ago = (x – 5) years.

His age 10 years later = (x + 10) years.

Using the given conditions, we get

(x – 5)(x + 10) = 16

x2 – 5x + 10x – 50 = 16

x2 + 5x – 66 = 0

x2 + 11x – 6x – 66 = 0

x(x + 11) – 6(x + 11) = 0

(x + 11)(x – 6) = 0

x = –11, or x = 6

But, x –11

Hence, Tanay's present age is 6 years.

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Q.32. The hypotenuse of a right triangle is

3 5 cm. If the smaller side is tripled and the larger

side is doubled, the new hypotenuse will be 15 cm.Find the length of each side. [2011 (T-II)]

Sol. Let the smaller side of the right triangle bex cm and the larger side be y cm.

Then, x2 + y2 = 2

3 5

[Using Pythagoras Theorem] x2 + y2 = 45 ......(i)If the smaller side is tripled and the larger side be

doubled, the new hypotenuse is 15 cm. (3x)2 + (2y)2 = 152

9x2 + 4y2 = 225 ...........(ii)From equation (i), we get y2 = 45 – x2

Putting y2 = 45 – x2 in equation (ii), we get9x2 + 4(45 – x2) = 225 5x2 + 180 = 225 5x2 = 45 x2 = 9 x = ± 3But, length of a side cannot be negative. Therefore,

x = 3.Putting x = 3 in (i), we get9 + y2 = 45 y2 = 36 y = 6

Hence, the length of the smaller side is 3 cm andthe length of the larger side is 6 cm.

Q.33. A takes 6 days less than the time takenby B to finish a piece of work. If both A and Btogether can finish it in 4 days, find the time takenby B to finish the work. [2011 (T-II)]

Sol. Suppose B alone takes x days to finish the work.Then, A alone can finish it in (x – 6) days.Now, (A's one day's work) + (B's one day's work)

=1 1

6x x

and, (A + B)'s one day's work =

1

4

1 1 1

6 4x x

6 1

( 6) 4

x x

x x

2

2 6 1

46

x

x x

8x – 24 = x2 – 6x

x2 – 14x + 24 = 0 x2 – 12x – 2x + 24 = 0

(x – 12)(x – 2) = 0

x = 12 or, x = 2

But, x cannot be less than 6. So, x = 12Hence, B alone can finish the work in 12 days.

PRACTICE EXERCISE 4.5 A

1. The sum of the squares of two consecutiveodd positive integers is 290. Find them.

2. The sum of two numbers is 8 and 15 timesthe sum of their reciprocals is also 8. Find thenumbers.

3. The product of two successive integralmultiples of 5 is 300. Determine the multiples.

4. The sum of the squares of two numbers is233 and one of the numbers is 3 less than twice theother number. Find the numbers.

5. Find two consecutive positive integers, thesum of whose squares is 365.

6. The difference of two numbers is 8 and thesum of their squares is 274. Find the numbers.

7. Find three consecutive positive numbers suchthat the square of the middle number exceeds thedifference of the squares of the other two by 60.

8. If a number is added to three times its

reciprocal, the result is3

55

. Find the number..

9. The sum of two natural numbers is 8.Determine the numbers if the sum of their

reciprocals is8

15. [Imp.]

10. The length of a rectangle is 3 cm more thanits width and its area is 40 cm2. Find the dimensionsof the rectangle.

11. One side of a rectangle exceeds its otherside by 2 cm. If its area is 195 cm2, determine thesides of the rectangle.

12. The denominator of a fraction is 3 morethan its numerator. The sum of the fraction and its

reciprocal is9

210

. Find the fraction. [Imp.]

13. A two-digit number is 4 times the sum ofits digits and also equal to twice the product of itsdigits. Find the number.

14. A passenger train takes one hour less for ajourney of 150 km, if its speed is increased by

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B. FORMATIVE ASSESSMENT

Activity

Objective : To find solution of quadratic equation x2 + bx + c = 0 by completing the square.

Materials Required : Coloured paper, a pair of scissors, geometry box and fevistick.

Procedure :

Case I : Let us find solution of x2 + 6x + 8 = 0.

1. Make a square of dimension x × x (Here, x = 5 cm) as shownin figure 1. Here, area of square is equal to 1st term ofpolynomial x2 + bx + c, i.e., x2.

5 km/hr from its usual speed. Find the usual speedof the train. [Imp.]

15. The time taken by a person to cover 150km was 2.5 hrs more than the time taken in thereturn journey. If he returned at a speed of10 km/hr more than the speed of going, what wasthe speed per hour in each direction?

16. An aeroplane takes 1 hour less for a journeyof 1200 km if its speed is increased by100 km/hr from its usual speed. Find its usualspeed. [Imp.]

17. The perimeter of a rectangular field is82 m and its area is 400 m2. Find the breadth of therectangle.

18. A dealer sells an article for Rs 24 and gainsas much per cent as the cost price of the article.Find the cost price of the article.

19. The sum of the ages of a man and his sonis 45 years. Five years ago, the product of theirages was four times the man's age at that time. Findtheir present ages. [Imp.]

20. The product of Shikha's age five years agoand her age 8 years later is 30, her age at both timesbeing given in years. Find her present age.

21. If two pipes function simultaneously, areservoir will be filled in 12 hours. One pipe fillsthe reservoir 10 hours faster than the other. Howmany hours will the second pipe take to fill thereservoir?

22. The sum of the squares of two positiveintegers is 208. If the square of the larger numberis 18 times the smaller, find the numbers.

23. A passenger train takes 2 hours less for ajourney of 300 km, if its speed is increased by5 km/hr from its usual speed. Find the usual speedof the train. [Imp.]

24. A pole has to be erected at a point on theboundary of a circular park of diameter 13 metresin such a way that the difference of its distancesfrom two diametrically opposite fixed gates A andB on the boundary is 7 metres. Is it possible to doso? If yes, at what distances from the two gatesshould the pole be erected? [V. Imp.]

25. A sailor can row a boat 8 km downstreamand return back to the starting point in 1 hour40 minutes. If the speed of the stream is 2 km/hr,find the speed of the boat in still water.

26. The length of a rectangle is thrice as longas the side of a square. The side of the square is4 cm more than the width of the rectangle. If theirareas being equal, find their dimensions.

27. A farmer prepares a rectangular vegetablegarden of area 180 sq. metres. With 39 metres ofbarbed wire, he can fence the three sides of thegarden, leaving one of the longer sides unfenced.Find the dimensions of the garden.

28. Out of a group of swans, 7/2 times thesquare root of the total number are playing on theshore of a pond. The two remaining ones areswinging in water. Find the total number of swans.

[V. Imp.]

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45

2. Add two strips of dimension 3 × x to figure 1 (2nd term of polynomial is +ve) to obtain figure

2. Here, area of each strip is equal to1

2of 2nd term of polynomial x2 + bx + c, i.e.,

1

2× bx.

3. Add and subtract a square of dimension 3 × 3 to figure 2, we will get an arrangement as shownin figure 3.

4. From above three figures we observe that

x2 + 6x = x2 + 3x + 3x + 9 – 9 = (x + 3)2 – 9

⇒ x2 + 6x + 8 = (x + 3)2 – 9 + 8 = (x + 3)2 – 1

∴ x2 + 6x + 8 = 0 ⇒ (x + 3)2 – 1 = 0

⇒ (x + 3)2 = 1 ⇒ x + 3 = + 1

⇒ x = – 4, –2.

Case II : Let us find solution of x2 – 6x + 8 = 0.

1. Make a grid of dimension x × x [x = 10] as shown infigure 4.

2. Cut out 60 strips (exactly shown in figure 5)from grid. [2nd term of polynomial is –ve]

Area of 60 strips = Area of two rectangles ofdimension 3 × x.

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[ Here, 3x =1

2of 6x, i.e., 2nd term of

square polynomial]

3. From figure 5, we observe that

Area of unshaded portion

= Area of square ABCD – Area of square EFGD

= (x – 3)2 – 32

x2 – 6x = (x – 3)2 – 9

x2 – 6x + 8 = (x – 3)2 – 9 + 8 = (x – 3)2 – 1

x2 – 6x + 8 = 0 ⇒ (x – 3)2 – 1 = 0

(x – 3)2 = 1 ⇒ x – 3 = + 1

x = 4, 2.

ANSWERS

A. SUMMATIVE ASSESSMENT

Practice Exercise 4.1A

1. (c) 2. (d) 3. (c) 4. (a)

5. (i), (iv), (vi), (vii), (x), (xi) 6. 3x = is a solution and x = 2 3− is not a solution.

7.3 4

and both are solutions2 3

x x= = − 8.1

3 and2

x x= = − are solutions

9. 4, 5 a b 10. (i)1

6k = − (ii) 22k a=


1. 2,1 2. 3,1 3. ,2 2

a b a b4.

3, 4

ba

a5.

2

21,

b

a

6.n mn

m

±− 7.

2

1,

a b

a b8.

1,a

a9. , 1 a a 10.

98 133,

33 33

11.2 6 6

,3 8

12.

43,

313. 1 14.

2,

a b

ab a b15. 0, a + b

16. –2, –3

217. 2, –1 18. 2, 3 19.

1, a

a20. ,

bc bc

ad ad

21.2 3 3

,3 4

22. 2 2

1 1,

a b23.

110,

5 24.

23, 1

5 25.

2 2

,2 2

a b

26.13 7

7,7

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1.1

, 2 22

2.7

3,3

3. 2, 1 4. 2, 3 2 5.

2,

b b

a a

6.1 17 1 17

,4 4 4 4

7.6 20

0,15

a b8. 6, 2 9.

2 2

2

− ±

10. no real roots 11. no real roots 12. 6a; a 13.3 10

7

±14.

4, 2

3

a

15.4

;b b

a a16. ;

2

pqpq 17.

3 5;

2 3 18.

5 5 4 5

2

− ± +

19. 2a ; – 3 20. 2p; q 21. 2 2

1 1;

a b22. ;

2 2

a b a b

23.

2( ) ( ) 4 ( )

2( )

c b b c a a b c

a b

− ± + + − −−

24. ( 1); ( 1) m n m n


1. (c) 2. (a) 3. (a) 4. (d) 5. (a)6. (b) 7. (b) 8. (b) 9. (a)

10. (i) 0 (ii) 12 (iii) 16 (iv) 12a2 (v) 9 8 2−(vi) 16 – 4b (vii) (2a – b)2 (viii) 1

11. (i) real and distinct (ii) real and distinct (iii) real and equal(iv) no real roots (v) real and distinct (vi) real and equal

(vii) no real roots (viii) real and distinct (ix) real and equal(x) no real roots.

12. (i) 8 (ii)1

2,2

(iii)1

,12

(iv) 5, –3 (v)1

3

(vi) 4, 3 (vii) 6, 1

5 (viii) 0, 3


1. 11, 13 2. 3, 5 3. 15, 20 or –20, –15 4. 8, 13

5. 13, 14 6. 15, 7 7. 9, 10, 11 8. 5 9. 3, 5

10. 5 cm, 8 cm 11. 13 cm, 15 cm 12.2

513. 36 14. 25 km/hr

15. 20 km/hr, 30 km/hr 16. 300 km/hr 17. 16 m 18. Rs 20

19. 36 yrs, 9 yrs. 20. 7 yrs 21. 30 hrs 22. 8, 12 23. 25 km/hr

24. At a distance of 5 metres from the gate B 25. 10 km/hr

26. 6 cm (l = 18 cm, b = 2 cm) 27. l = 15 m, b = 12 m 28. 16

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