3. QUADRATIC EQUATION - TopperLearning

www.plancess.com

3. QUADRATIC EQUATION

1. IntroductionWhen a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x) is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t + 3

5 = 0

etc. are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax2 + bx + c, a 0, then f(x) = 0 i.e., ax2 + bx + c = 0, a 0 is called a quadratic equation. Such equations arise in many real life situations. In this chapter, we will learn about quadratic equations and various ways of �nding their zeroes or roots. In the end of the chapter, we will also discuss some applications of quadratic equations in daily life situations.

2. Quadratic EquationA polynomial equation of degree two is called a quadratic equation.

e.g. 2x2 – 3x + 1 = 0, 4x – 3x2 = 0 and 1 – x2 = 0

General form of quadratic equations: ax2 + bx + c = 0, where a, b, c, are real numbers and a 0.

Moreover, it is general form of a quadratic equation in standard form.

Roots of quadratic equation: x = α is said to be root of the quadratic equation ax2 + bx + c = 0, a 0 if x = α satis�es the quadratic equation i.e. in other words the value of a 2α + bα + c is zero.

Solving a quadratic equation: �e determination of all the roots of a quadratic equation is called solving the quadratic equation.

Illustration 1: In each of the following, determine whether the given values are the solution of the given equation or not:

(i) 2

2 5–

xx+2 = 0 ; x = 5 , x = 1

2

(ii) 2 2 2a x 3abx 2b 0 ; x = a b, x

b a

Solution: (i) Putting x = 5 in the given equation, we get

3.2 Quadratic Equation

www.plancess.com

2

2 5– 2

5(5)

i.e. 225

- 1 + 2

i.e. 225

+ 1

i.e., x = 5 does not satisfy

Putting x = 12

in the given equation, we get

2

2 5– 2

1122

2 5– 2

1 14 2

=8 – 10 + 2=0

i.e. x = 12

satis�es the given equation.

Hence, x = 5 is not a solution but x = 12

is a solution of2

2 5– 2 0

xx.

(ii) Putting x = ab

in the given equation.

22 2a a

a – 3ab 2bb b

And 2

2 2b ba – 3ab 2b

a a

⇒4

2

a

b2 22b 3a And 0

i.e., x = ab

does not satisfy

Putting x = ba

in the given equation.

22 2b b

a – 3ab 2ba a

=0

i.e. x = ba

satis�es the given equation.

Hence x = ba

is a solution but x = ab

is not a solution of a2x2 – 3abx + 2b2 = 0.

Illustration 2: Find the values of p and q for which x = 34

and x = – 2 are the roots of the equation

px2 + qx – 6 = 0.

3.3 Foundation for Mathematics

www.plancess.com

Solution: Since x = 34

and x = – 2 are the roots of the equation px2 + qx – 6 = 0.

2

3 3p q

4 4 – 6 = 0 and p(–2)2 + q(–2) – 6 = 0

⇒ p × 916

+ q × 34

– 6 = 0 and 4p – 2q – 6 = 0

⇒ 9p 12q 9616

= 0 and 4p – 2q – 6 = 0

⇒ 9p + 12q – 96 = 0 and 4p – 2q – 6 = 0 ⇒ 3p + 4q – 32 = 0 ...(1) and 2p – q – 3 = 0 ..(2) Multiplying (2) by 4, we get 8p – 4q – 12 = 0 ...(3) Adding (1) and (3), we get p = 4 Putting the value of p in equation (2), we get 2 × 4 – q – 3 = 0 ⇒ q = 5 Hence, p = 4, q = 5.

3. Methods of Solving Quadratic Equations

3.1 By Factorisation MethodStep-I: Factorize the constant term of the given quadratic equation.

Step-II: Express the coe�cient of middle term as the sum or di�erence of the factors obtained in step I (in such a way that, the product of these two factors will be equal to the product of the coe�cient of x2 and constant term).

Step-III: Split the middle term in two parts obtained in step-II

Step-IV: Factorize the quadratic equation obtained in step-III by grouping method.

Illustration 3: Solve the quadratic equation 5x2 = – 16x – 12 by factorisation method.

Solution: 25x 16x 12

or, 25x 16x 12 0

or, 25x 10x 6x 12 0

or, 5x(x 2) 6(x 2) 0

or, (x 2) (5x 6) 0

i.e. either x 2 0 x 2⇒ or 65x 6 0 x

5⇒

3.2 By Completing the Square Step-I: Obtain the quadratic equation. Let the quadratic equation be ax2 + bx + c = 0, a 0.

Step-II: If coe�cient of x2 is not unity then make it unity by dividing throughout by a.


www.plancess.com

i.e. x2 + ba x + c

a = 0

Step-III: Shift the constant term ca on R.H.S. to get x2 + b

a x = – ca

Step-IV: Add square of half of the coe�cient of x. i.e. 2

b2a

on both sides to obtain

x2 + 2 b2a

x +2

b2a

= – ca +

2b2a

Step-V: Write L.H.S. as the perfect square and simplify R.H.S. to get 2

bx

2a=

2

2

b – 4ac

4a

Step-VI: Take square root of both sides to get bx

2a= ±

2

2

b – 4ac

4a

Step-VII: Obtain the values of x by shifting the constant term b2a

on R.H.S. i.e. x = – b2a

±2

2

b – 4ac

4a

Illustration 4: Solve the equation x2 – ( 3 + 1) x + 3 = 0 by the method of completing the square.

Solution: We have, 2x 3 1 x 3 0

2x 3 1 x 3

2x 2⇒ 3 1

2 x +

23 12

= 3 + 2

3 12

⇒

23 1x –2

= 2– 4 ( 3 1)3

4

⇒

23 1x –2

= 2

–3 12

⇒ 3 1x –2

–3 12

⇒ 3 1x2

–3 12

x 3, 1⇒

Hence, the roots are 3 and 1.

3.3 By Quadratic Formula “Sreedharacharya’s Rule”

Consider quadratic equation ax2 + bx + c = 0, a 0 then x =2–b b – 4ac

2a

�e roots of x are 2–b b – 4ac

2a and

2–b – b – 4ac2a

⇒x = –b D2a

or, x = –b – D2a

, where D = b2 – 4ac


www.plancess.com

�us, if D = b2 – 4ac 0, then the quadratic equation ax2 + bx + c = 0 has real roots α and βgiven by

α = –b D2a

and β = –b – D2a

3.3.1 Discriminant

If ax2 + bx + c = 0, a 0 (a, b, c R) is a quadratic equation, then the expression b2 – 4ac is known as its discriminant and is generally denoted by D or .

PLANCESS CONCEPTS

If the sum of the co-e�cient of a quadratic equation ax2 + bx + c = 0, a 0 (a, b, c R) i.e. a+b+c=0, is equal to zero, then x = 1 is a root of the equation.

Aman Goel

Gold Medalist, INPhO

4. Nature and Relation of the Roots with the Coefficients4.1 Nature of the RootsLet the quadratic equation be ax2 + bx + c = 0 ... (i)

Where a 0 and a, b, c Q.

�e roots of the given equation are given by x = –b D2a

, where D = b2 – 4ac is the discriminant.

i.e., if α and β are two roots of the quadratic equation (i). �en, α = –b D2a

and β = –b – D

2a

Now, the following cases are possible.

Case-I: When D > 0, Roots are real and unequal (distinct).

�e roots are given by α = –b D2a

and β = –b – D2a

Case-II: When D = 0, Roots are real and equal and each root = –b2a

Case-III: When D < 0, No real roots exist. Both the roots are imaginary.


www.plancess.com

Discriminant

If D 0 then If D < 0, then

�e equation has real roots the equation has no real root.

If D 0, then the equation has If D = 0, then the equation has

two real, distinct roots two real and equal roots i.e. b2a

α β

If D is a perfect square, then the equation If D is not a perfect square, then the equation

has two real, distinct and rational roots. has two real, distinct and irrational roots.

PLANCESS CONCEPTS

Consider a quadratic equation ax2 + bx + c = 0, where a, b, c Q, a 0 and D > 0 then:

(i) If D is a perfect square, then roots are rational and unequal.

(ii) If D is not a perfect square, then roots are irrational and unequal. If one root is of the form p + q (where p is rational and q is a surd) then the other root will be p – q .

Shivam Agarwal


4.2 Sum and Product of the RootsLet α and β be the roots of the quadratic equation ax2 + bx + c = 0, a 0.

�en α = 2–b b – 4ac

2a and β =

2–b – b – 4ac2a

�e sum of roots α β = – ba = –

2

Coeff. of x

Coeff.of x

and, product of roots = .α β = ca =

2

constant term

coefficient of x


www.plancess.com

PLANCESS CONCEPTS

Formation of Quadratic Equation

�e quadratic equation whose roots are α and β is given by x2 – (α β ) x + α β = 0

i.e. x2 – (sum of the roots) x + product of the roots = 0

Vaibhav Gupta


4.3 Relation between Roots and CoefficientIf roots of quadratic equation ax2 + bx + c = 0 (a 0) are α and β then:

(i) 2( ) ( ) 4α β α β αβ 2b 4ac D

a a

(ii) 2

2 2 22

b 2ac( ) 2

aα β α β αβ

(iii) 3 3 3( ) 3 ( )α β α β αβ α β 2

3

b(b 3ac)

a

(iv) 4 4 2 2 2 2[( ) 2 ] 2α β α β αβ α β 22 2

2 2

b 2ac c2

a a

(v) 5 5 3 3 2 2 2 2( )( ) ( )α β α β α β α β α β 3 2 2 2[( ) 3 ( )] [( ) 2 ] [ ( )]α β αβ α β α β αβ α β α β

(vi) 2 2 2( ) ( ) 4α β α β α β αβ2

2 2

b Db b 4ac

a a

(vii) 3 3 3( ) 3 ( )α β α β αβ α β 22( ) 4α β αβ α β αβ 2 2

3

(b ac) b 4ac

a

(viii) 4 4 2 2 2 2( )( )α β α β α β 2 2

4

b(b 2ac) b 4ac

a

(ix) 2 2 2( )α αβ β α β αβ

(x) 2 2 2( ) 2α β α β α β αβ

β α αβ αβ

(xi) 2 2 ( )α β β α αβ α β .

(xii) 2 2 4 4 2 2 2 2 2

2 2 2 2

( ) 2α β α β α β α ββ α α β α β

Illustration 5: Find the nature of the roots of the following equations. If the real roots exist, �nd them.

(i) 2x2 – 6x + 3 = 0 (ii) 2x2 – 3x + 5 = 0


www.plancess.com

Solution: (i) �e given equation 2x2 – 6x + 3 = 0

Comparing it with ax2 + bx + c = 0, we get a = 2, b = – 6 and c = 3.

∴ Discriminant, D = b2 – 4ac = (–6)2 – 4×2×3 = 36 – 24 = 12 > 0

D > 0, roots are real and unequal.

Now, by quadratic formula, x = –b D2a

= 6 122 2

= 6 2 34

= 3 32

Hence the roots are x = 3 32

, 3 – 32

(ii) Here, the given equation is 2x2 – 3x + 5 = 0;

Comparing it with ax2 + bx + c = 0, we get a = 2, b = – 3 and c = 5

Discriminant, D = b2 – 4ac = 9 – 4×2×5 = 9 – 40 = – 31

D < 0, the equation has no real roots.

5. Equation in Terms of the Roots of another EquationIf ,α β are roots of the equation ax2 + bx + c = 0 then equation whose roots are

(i) 2, ax bx c 0α β⇒ (Replace x by – x)

(ii) 21 1, cx bx a 0⇒

α β 1(Replace x by )

x

(iii) n n 1/n 2 1/n, ;n N a(x ) b(x ) c 0α β ⇒ 1/n(Replace x by x )

(iv) 2 2k ,k ;ax kbx k c 0α β⇒ (Replace x by x/k)

(v) 2k ,k a(x k) b(x k) c 0α β⇒ (Replace x by x-k)

(vi) 2 2, k ax kbx c 0k kα β

⇒ (Replace x by kx)

(vii) 1/n 1/n n 2 n, ;n N a(x ) b(x ) c 0α β ⇒ n(Replace x by x )

6. Different Types of EquationsEquations which are not quadratic at a glance but can be reduced to quadratic equations by suitable transformations. Some of the common types are:

Type-I: ax4 + bx2 + c = 0

�is can be reduced to a quadratic equation by substituting x2 = y i.e., ay2 + by + c = 0

E.g. Solve 2x4 – 5x2 + 3 = 0

Putting x2 = y, we get 2y2 – 5y + 3 = 0

⇒ (2y – 3) (y – 1) = 0 ⇒ y = 32

or 1 ⇒x2 = 32

or 1 ⇒ x = ± 32

or ± 1


www.plancess.com

Type-II: a {p(x)}2 + b⋅p (x) + c = 0 where p(x) is an expression in ‘x’

Put p(x) = y, to get the quadratic equation ay2 + by + c = 0.

e.g. Solve (x2 + 3x)2 – (x2 + 3x) – 6 = 0, x R

Putting x2 + 3x = y, we get y2 – y – 6 = 0

Solving, we get y = 3 or – 2

⇒ x2 + 3x = 3 or x2 + 3x = – 2

⇒x = –3 212

or x = – 2 or – 1.

Type-III: ap(x) + bp(x)

= c, where p(x) is an expression in x.

Put p(x) = y to obtain the quadratic equation ay2 – cy + b = 0

e.g. Solve x x 1 34x 1 x 15

( x 1 and 0 )

Putting xx 1

= y, we get, y + 1 34y 15

⇒ 15y2 – 34y + 15 = 0 ⇒ y = 5 3or

3 5

⇒ xx 1

= 53

or 53

= xx 1

⇒ x = –5 3or

2 2

Type-IV: (i) a 22

1 1x + + b x + + c = 0xx

(ii) a 22

1 1x + + b x – + c = 0xx

If the coe�cient of b in the given equation contains x + 1x

, then replace 22

1x

xby

22

2

1x

x – 2 and

put x+ 1x

= y. In case the coe�cient of b is x – 1x

, then replace 22

1x

xby

21

x –x

+ 2 and put x – 1x

= y.

e.g. Solve 9 22

1 1x – 9 x – 52 0

xx

Putting x + 1x

= y, we get : 9(y2 – 2) – 9y – 52 = 0

⇒ y = 103

or y = – 73

⇒ x + 1x

= 103

or x + 1x

= – 73

,

⇒ x = 13

or 3 or x = –7 136

.

Type-V: (x + a) (x + b) (x + c) (x + d) = k, such that a + b = c + d.

Rewrite the equation in the form {(x + a) (x + b)} · {(x + c) (x + d)} = k

Put x2 + x(a + b) = x2 + x(c + d) = y to obtain a quadratic equation in y i.e. (y + ab) (y + cd) = k.


www.plancess.com

e.g. Solve (x + 1) (x + 2) (x + 3) (x + 4) = 120

∵1 + 4 = 2 + 3, we write the equation in the following form:

{(x + 1) (x + 4)} {x + 2) (x + 3)} = 120

⇒ (x2 + 5x + 4) (x2 + 5x + 6) = 120

Putting x2 + 5x = y, we get (y + 4) (y + 6) = 120

⇒ y = – 16 or 6

⇒ x2 + 5x = – 16 or x2 + 5x = 6

⇒ x = – 6 or 1 (x2 + 5x + 16 has no real solution)

Type-VI: ax + b ± cx + d = e

Square both sides and simplify in such a manner that the expression involving radical sign on one side and all other terms are on the other side. Square both sides of the equation thus obtained and simplify it to obtain a quadratic in x. Reject these values which do not satisfy ax + b 0 and cx + d 0.

e.g. Solve : 4 x x 9 5

⇒ 4 x = 5 – x 9

⇒ x + 15 = 5 x 9 (on squaring both sides)

⇒ (x + 15)2 = 25 (x + 9) (on squaring both sides)

⇒ x = 0 or – 5

Clearly, x = 0 and x = – 5 satisfy 4 – x 0 and x + 9 0.

Hence, the roots are 0 and – 5.

NOTE: Always try to substitute the �nal values in the given equation and check whether it satis�es or not.

Illustration 6: �e sum of the squares of two consecutive odd positive integers is 290. Find the integers.

Solution: Let the two consecutive odd positive integers be x and (x + 2). �en, x2 + (x + 2)2 = 290 ⇒ x2 + x2 + 4x + 4 = 290 ⇒ x2 + 2x – 143 = 0 ⇒x2 + 13x – 11x – 143 = 0 ⇒ x (x + 13) – 11 (x + 13) = 0

⇒ (x + 13) (x – 11) = 0 ⇒ x = – 13 or x = 11 But – 13, is not an odd positive integer. Hence, the required integers are 11 and 13.


www.plancess.com

7. Roots of Cubic and Quadratic EquationsIf the equation is of a cubic form, ax3 + bx2 + cx +d = 0, and ,α β , are the 3 roots of the cubic equation, then

ba

α β

ca

αβ β α

da

αβ

If the equation is of biquadratic form, ax4 + bx3 + cx2 + dx + e = 0 and , ,α β and are the four roots of the quadratic equation.

ba

α β

ca

αβ α α β β

da

αβ αβ α β

ea

αβ

Illustration 7: Seven years ago Varun’s age was �ve times the square of Swati’s age. �ree years hence, Swati’s age will be two �fth of Varun’s age. Find their present ages.

Solution: Let the present ages of Varun and Swati be x years and y years respectively.

Seven years ago, Varun’s age = (x – 7) years and Swati’s age = (y – 7) years.

(x – 7) = 5 (y – 7)2 ⇒ x – 7 = 5 (y2 – 14 y + 49)

⇒ x = 5y2 – 70y + 245 + 7 ⇒ x = 5y2 – 70y + 252 ... (i)

�ree years hence, Varun’s age = (x + 3) years and Swati’s age = (y + 3) years.

(y + 3) = 25

(x + 3) ⇒ 5y + 15 = 2x + 6 ⇒ x = 5y 92

... (ii)

From (i) and (ii) we get 5y2 – 70y + 252 = 5y 92

⇒ 10y2 – 140y + 504 = 5y + 9 ⇒ 10y2 – 145y + 495 = 0 ⇒ 2y2 – 29y + 99 = 0 ⇒ 2y2 – 18y – 11y + 99 = 0 ⇒ 2y (y – 9) – 11 (y – 9) = 0


www.plancess.com

⇒ (y – 9) (2y – 11) = 0

⇒ y = 9 or y = 112

y = 112

is not possible 117

2

So, y = 9.

x = 5 9 92

= 27 [From (ii)]

Hence, the Varun’s present age is 27 years and Swati’s present age is 9 years.

Illustration 8: Is it possible to design a rectangular park of perimeter 80 cm and area 400 m2? If so, �nd its length and breadth.

Solution: Let the length and breadth of the rectangular park be l and b respectively. �en, 2 (l + b) = 80 ⇒ l + b = 40 ⇒ l = (40 – b) And area of the park = 400 m2 lb = 400 m2

⇒ (40 – b) b = 400 ⇒ 40b – b2 = 400 ⇒ b2 – 40b + 400 = 0 ⇒ (b – 20)2 = 0 ⇒ b – 20 = 0 ⇒ b = 20 m

l = 40 – b = 40 – 20 = 20 m Hence, length and breadth of the park are 20 m and 20 m respectively. �us, it is possible to design a rectangular park of perimeter 80 m and area 400 m2

Illustration 9: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution: Let the speed of the train be x km/h. �en,

Time taken to cover the distance of 360 km = 360x

hours.

If the speed of the train increased by 5 km/h. �en,

Time taken to cover the same distance = 360x 5

hours

According to the question, 360x

– 360x 5

= 1

⇒ 360(x 5) – 360xx(x 5)

= 1 ⇒ 360x + 1800 – 360x = x2 + 5x

⇒ x2 + 5x – 1800 = 0 ⇒ x2 + 45x – 40x – 1800 = 0


www.plancess.com

⇒x(x + 45) – 40 (x + 45) = 0 ⇒ (x + 45) (x – 40) = 0

⇒ x = – 45 or x = 40

But the speed cannot be negative. Hence, the speed of the train is 40 km/h.

Illustration 10: 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.

Solution: Let the number of students be x. �en,

�e number of apples received by each student = 300x

If there is 10 more students, i.e., (x + 10) students.

�en, the number of apples received by each student= 300x 10

According to the question, 300x

– 300x 10

= 1

⇒ 300x 3000 – 300xx(x 10)

= 1 ⇒ 3000 = x2 + 10x

⇒ x2 + 10x – 3000 = 0 ⇒ x2 + 60x – 50x – 3000 = 0

⇒ x(x + 60) – 50 (x + 60) = 0 ⇒ (x + 60) (x – 50) = 0

⇒ x = – 60 or x = 50

But the number of students cannot be negative. Hence, the number of students is 50.

Illustration 11: If the sum of the roots of ax2 + bx + c = 0 is equal to the sum of the squares of the roots then �nd the condition.

Solution: If ,α β are the roots of ax2 + bx + c = 0 then

α β= –ba

, .α β = ca

Given that 2 2α β α β

⇒–ba

2b

–a

c2.

a

2

2

b

a

2ca

⇒–ba

=2

2

b – 2ac

a ⇒ 2 2–b

a b 2aca

⇒ 2ab b 2ac 22ac b ab⇒

22ac b ab Is the required condition.

Illustration 12: If α and β are the roots of quadratic equations ax2 – bx + c = 0, form the equation

whose roots are 2

β

α and

2

α

β


www.plancess.com

Solution: α and β are the roots of ax2 – bx + c = 0. �en α β = ba

and .αβ = ca

We want equation with roots 2

β

α and

2

α

β,

Sum of the roots ‘S’ = 2

β

α +

2

α

β

3 3

2 2

β α

α β =

3

2

( ) – 3 ( )

( )

α β αβ α β

αβ ….. {As a3 + b3 = (a + b)3 – 3ab (a+b)}

3

2

b c b– 3 .

a a a

ca

=

3

3 2

2

2

b 3cb–

a ac

a

3

3

b – 3abc

a ×

2

2

a

c =

3

2

b – 3abc

c a

Product of roots P =2

β

α.

2

α

β = 1

αβ = 1

c/ a = a

c

�e required equation is x2 – (S)x + P = 0 or x2 – 3

2

b – 3abc

c a x + a

c = 0

or, c2ax2 – (b3 – 3abc)x + a2c = 0

8. Geometrical Representation of Quadratic Expression Consider the quadratic expression, y = ax2 + bx + c, a 0 and a, b, c R then:

(i) �e graph between x,y is always a parabola. If a > 0, then the shape of the parabola is concave upwards and if a < 0 then the shape of the parabola is concave downwards.

(ii) �e graph of y = ax2 + bx + c can be divided into 6 categories which are as follows:

(Let the roots of the equation a x2 + bx + c = 0 be α and β )

Remark :

(i) �e quadratic expression ax2 + bx + c > 0 x R ⇒ a > 0, D < 0 (�gure 3.1)

Figure 3.1


www.plancess.com

(ii) �e quadratic expression ax2 + bx + c < 0 x R ⇒ a < 0, D < 0 (�gure 3.2)

Figure 3.2


www.plancess.com

SUMMARY

• If f(x) is a quadratic polynomial i.e., f(x) = ax2 + bx + c, a 0, then f(x) = 0 i.e., ax2+ bx + c = 0, a 0 is called a quadratic equation.

• Roots of quadratic equation: x = α is said to be root of the quadratic equation ax2 + bx + c = 0, a 0 if a 2α + bα + c = zero.

• Methods of Solving Quadratic Equations

(i) Factorisation method

(ii) Completing the square method

(iii) By Quadratic Formula “Sreedharacharya’s Rule”

• Discriminant: If ax2 + bx + c = 0, a 0 (a, b, c R) is a quadratic equation, then the expression b2 – 4ac is known as its discriminant and is generally denoted by D or .

• Nature of the Roots of the Quadratic Equation:

• Consider a quadratic equation a x2 + bx + c = 0, where a, b, c Q, a 0

• Case-I: When D > 0, Roots are real and unequal (distinct).

(i) If D is a perfect square, then roots are rational and unequal.

(ii) If D is not a perfect square, then roots are irrational and unequal. If one root is of the form p +q then, the other root is given by p - q

• Case-II: When D = 0, Roots are real and equal

• Case-III: When D < 0, No real roots exist. Both the roots are imaginary.

• Sum of roots = – 2

Coeff. of x

Coeff.of x

• Product of roots = 2

constant term

coefficient of x

• Equation in terms of the roots of another equation:

(i) 2, ax bx c 0α β⇒ (Replace x by – x)

(ii) 21 / ,1 / cx bx a 0α β⇒ (Replace x by 1/x)

(iii) n n 1/n 2 1/n, ;n N a(x ) b(x ) c 0α β ⇒ 1/n(Replace x by x )

(iv) 2 2k ,k ;ax kbx k c 0α β⇒ (Replace x by x/k)

(v) 2k ,k a(x k) b(x k) c 0α β⇒ (Replace x by x-k)

(vi) 2 2, k ax kbx c 0k kα β

⇒ (Replace x by kx)

(vii) 1/n 1/n n 2 n, ;n N a(x ) b(x ) c 0α β ⇒ n(Replace x by x )


www.plancess.com

SOLVED EXAMPLESExample 1. A man walks a distance of 48 km in a given time. If he walks 2 km an hour faster, he will perform the journey 4 hours before. Find his normal rate of walking.

Solution: Let the normal rate of walking of the man be x km/hour.

Time taken to walk 48 km= 48x

hours

At 2 km an hour faster he will take = 48x 2

hours.

�is time is 4 hour less than the usual time.

48 48

4x x 2

48 (x 2) 48x 4x (x 2)

2x 2x 24 0 ⇒ (x 4) (x 6) 0

x 4 Or 6

Rate of walking is 4 km/hour as it cannot take negative value.

Example 2. Evaluate 120

120

20....Solution: Let 1

x 20x

�erefore, 2x 20x 1 0

�is gives 20 404x 10 101

2 or 20 404

x 10 1012

Since, the given expression cannot be negative. �erefore, we neglect the negative value,

i.e.10 101 . Hence, the desired value of the expression is10 101 .

Example 3. Find the condition that the quadratic equations 2x ax b 0 and 2x bx a 0 may have a common root.

Solution: Let α be a common root of the given equations.

�en 2 a b 0α α and 2 b a 0α α

By the method of cross-multiplication, we get 2

2 2

1b a b aa b

α α

�is gives 2 2

2 a b(a b)

b aα and 1α

2(1) (a b) 1 a b⇒ ⇒

a b 1 0⇒ is the required condition.


www.plancess.com

Example 4. Solve 24x 4x 1 3 x

Solution: 2 24x 4x 1 3 x (2x 1) 3 x⇒

(2x 1) 3 x⇒

2x 1 3 x⇒ or (2x 1) 3 x

3x 2⇒ or 2x 1 x 3

2

x3

⇒ or x 4

Hence, 24 x

3

Example 5. If the roots of 2 2 2 2 2(q r ) x 2r (p q)x (r p ) 0 are equal, show that p, q and r in G.P i.e. 2r pq .

Solution: Since, the roots are equal

⇒ Discriminant = 0 2 2 2 2 22r(p q) 4(q r )(r p ) 0⇒

2 2 2 2 2 2 2 4 2 24r (p 2pq q ) 4 (q r q p r r p ) 0⇒

2 2 2 2 2 2 2 2 2 4 2 2(p r 2pqr q r ) (q r q p r r p ) 0⇒ 2 2 2 42pqr q p r 0⇒

4 2 2 2r 2pqr q p 0⇒

2 2 2(r pq) 0 r pq⇒ ⇒ ⇒ p. q. and r in G.P.

Example 6. If the roots of 2ax bx b 0 are in the ratio m:n, then �nd the value of m nn m

.

Solution: Let the roots bem , nα α .

bm n

aα α And b

m na

α α

We have to eliminateα .

b b(m n)

a a(m n)α ⇒ α

Now,

2

2

2 2

2

bmn

ab b

mnaa (m n)

mnb1

a(m n)

α

⇒

⇒


www.plancess.com

2

2

mn ab(m n)

Taking reciprocacal of both side, we get

(m n) bmn a

⇒

⇒

Taking square root on both side, we get

m n bamn

m n bamn mn

m n bn m a

⇒

⇒

⇒

Example 7. Solve for x 1 2x 1x; 3 3 270

Solution: x 1 2x 13 3 270

x 2x3.3 3 .3 270⇒ 2x x3 3 90⇒

Substituting 3x = a , we get,

a2 + a= 90

2a a 90 0⇒ 2a 10a 9a 90 0⇒

(a 10)(a 9) 0⇒

a 9⇒ or a 10

If x3 9 , then x 2

If x3 10 , which is not possible.

x 2

Example 8. Find the solution set of the equation 8x 5 7

x 5 ( x 5 )

Solution: 8x 5 7

x 5

Put u x 5

8

u 7u

Multiply both sides by u, we get,

2u 8 7u


www.plancess.com

i.e. 2u 7u 8 0

�e equation reduces to 2u 7u 8 0

i.e., (u 8) (u 1) 0 u 8 or u 1

u = x + 5

i.e. if x 5 8 x 8 5 3⇒

if x 5 1 x 1 5 6⇒

Roots are x = 3 and x = - 6

�e solution set = {-6, 3}.

Example 9. A length of 60 cm is divided into equal parts. What is the number of these parts if, when this number is increased by unity, the length of each part is decreased by 1 mm?

Solution: �e length = 60 cm or 600 mm. let n be the number of parts.

Hence, length of each part = 600n

mm

When length of each part is reduced by 1 mm, the new length of each part 6001 mm

n

When number of parts is increased by unity, the resulting number of parts (n 1)

2 2

6001 (n 1) 600

n

600600 n 1 600

n600 n n 0 n n 600 0

⇒

⇒ ⇒

2n 25 n 24 n 600 0⇒

i.e. n (n 25) 24 (n 25) 0

i.e. (n 24) (n 25) 0

n 24 Or n 25 (inadmissible since n cannot be negative)

Number of parts = 24


www.plancess.com

Example 10. An aeroplane travelled a distance of 400 km at an average speed of x km/h, on the return journey, the speed was increased by 40 km/h. Write down an expression for the time taken for:

(i) �e onward journey (ii) the return journey

If the return journey took 30 minutes less than the onward journey, write down an equation in x and �nd its value

Solution: (i) Time taken for onward journey, Distance travelled 400t

Speed x …. (i)

(ii) Time taken for return journey 400t

x 40 …(ii)

Now, according to question, we have; 400 400 1x x 40 2

130 min = h

2

800 x 40400x 2 (x 40)

⇒

2800x 32000 800 x x 40 x⇒

2x 40 x 32000 0⇒

2x 200 x 160 x 32000 0⇒

x (x 200) 160(x 200) 0⇒

(x 160)(x 200) 0⇒

Either x 160 0 or x 200 0

x 160⇒ or x 200

Since, speed cannot be negative, as it is a magnitude of velocity so x 200 is neglected.

Hence, x 160 km/hr.


www.plancess.com

EXERCISE 1 — For School Examinations

Fill in the BlanksDirections: Complete the following statements with an appropriate word/term to be �lled in the blank space(s).

Q1. A quadratic equation 2ax bx c 0 has two distinct real roots, if 2b 4ac .

Q2. Two consecutive positive integers, sum of whose squares is 365 are .

Q3. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. �e speed of the stream .

Q4. �e values of k for which the equation 22x kx x 8 0 will have real and equal roots are .

Q5. Ifα , β are roots of the equation 2ax bx c 0 , then the quadratic equation whose roots are a bα and a bβ is .

Q6. If r, s are roots of 2ax bx c 0 , then 2 2

1 1

r s is .

Q7. If α is one of the roots of a quadratic equation 2x 2px p 0 , then the other root is .

Q8. If the sum of the squares of the roots of 2x px 3 0 is 10, then the value of p= .

Q9. If ,α β are the roots of 2x bx c 0 and α + h , β+ h are the roots of 2x qx r 0 , then h= .

Q10. In a homogeneous expression, all the terms will have the .

Q11. A quadratic equation cannot have more than roots.

True / False

Directions: Read the following statements and write your answer as true or false.

Q12. A quadratic equation cannot be solved by the method of completing the square.

True False

Q13. (x 2)(x 1) (x 1) (x 3) is a quadratic equation.

True False

Q14. 2x x 306 0 Represent a quadratic equation where product of two consecutive positive integer is 306.

True False


www.plancess.com

Q15. �e value of x satisfying the equation 2 2 2x p (q x) is 2 2p q

2 True False

Q16. Sum of the reciprocals of the roots of the equation 2x px q 0 is 1/p.

True False

Q17. �e nature of roots of equation 2x 2x 3 3 0 are real and equal.

True False

Q18. If 2 2a b 0 , then a > b or a < - b.

True False

Match the Following Columns

Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II

Q19. Column II give pair of two numbers for solution to problems given in Column I, match them correctly.

Column I Column II

(A) �e sum of the squares two positive integers is 208. If the square of the larger number is 18 times the smaller.

(p) (7, 49)

(B) A year ago, the father was eight times as old as his son. Now his age is the square of his son’s age.

(q) (5, 29)

(C) �e age of father is equal to the square of the age of his son. �e sum of the age of father and �ve times the age of the son is 66 years.

(r) (36, 6)

(D) Two years ago, Jacob’s age was three times the square of John’s age. In three year’s time. John’s age will be one-fourth of Jacob’s age.

(s) (8, 12)

Very Short Answer Questions

Directions: Give answer in one word or one sentence.

Q20. If the sum of the roots of the equation is 2 and sum of their cubes is 98, then �nd the equation.

Q21. Solve for x 1 xx : (3 6) (3 3) 1


www.plancess.com

Q22. If (cos 30 + sin 30) is a root of the quadratic equation then, �nd the quadratic equation.

Q23. For what value of p will the equations have real roots?2px 3x 4 0

Q24. Find the inequality, 2

2

3x 7x 82

x 1

Q25. Solve: 29x 6x 1 5 x

Q26. Find the value(s) of k for which each of the following quadratic equation has equal roots:22x kx 3 0

Q27. �ree consecutive natural numbers are such that the square of the middle number exceeds the di�erence of the square of the other two by 60, �nd the numbers.

Short Answer Questions

Directions: Give answer in two to three sentences.

Q28. Evaluate 6 6 6 ......

Q29. Solve: 1 1 1 1x 5 x 4 x 2 x 7

( x 5, 4, 2 and 7 )

Q30. �e sum of the ages of a father and his son is 45 years. Five years ago, the product of their age (in years) was 124. Determine their present ages.

Q31. A segment AB of 2m length is divided at C into two parts such that 2AC AB . CB . Find the length of the part CB.

Q32. Two persons while solving a quadratic equation, committed the following mistakes:

One of them made a mistake in the constant term and got the roots as 5 and 9.

Another one committed an error in the coe�cient of x and he got the roots as 12 and 4.

But in the meantime, they realized that they are wrong and they managed to get it right jointly. Find the quadratic equation.

Q33. Solve the equation: (x + 1) (x + 2) (x + 3) (x + 4) – 8 = 0

Q34. Determine the value(s) of p for which the quadratic equation 24x 3px 9 0 has real roots.

Q35. If the roots of the equation 2(a b)x (b c)x (c a) 0 are equal , prove that 2a = b + c.

Q36. Solve for 1 1 1 1x : ,a 0,b 0 ,x 0

a b x a b x


www.plancess.com

Q37. Two water taps together can �ll a tank in 39

8 hours. �e tap of larger diameter takes 10 hours

less than the smaller one to �ll the tank separately. Find the time in which each tap can �ll the tank separately.

Q38. Find the value of k if 2x x k 0 and 2x 10x (2k 3) 0 have 3 as a common root.

Q39. Solve the equation 2x px 45 0 , it is being given that the squared di�erence of its roots is equal to 144.

Q40. A number consists of two digits whose product is 18. If 27 is added to the number, the number formed will have the digits in reverse order, when compared to the original number. Find the number. Find the number.

Long Answer Questions

Directions: Give answer in four to �ve sentences.

Q41. A group of girls planned a picnic. �e budget for food was ₹ 2400. Due to illness, 10 girls could not go to the picnic and cost of food for each girl increased by ₹ 8. How many girls had planned the picnic?

Q42. A plane left 40 minutes late due to bad weather and in order to reach the destination 1600 km away in time, it had to increase its speed by 400 km/hour from its usual speed. Find its usual speed.

Q43. A takes 12 days less than B to �nish a piece of work. If A and B together can �nish the work in 8 days, �nd the time taken by B to �nish the work.

Q44. If ,α β are the roots of the equation 2ax bx c 0 , then �nd the equation whose roots are 1

αβ

and 1β

α.

Q45. Find the solution of equation p q x q r x r p x 3x

0r p q p q r

(r 0, p 0 and q 0)

Q46. If α and β are roots of the equation2 2 2 2A(x m ) Amx cm x 0 , then �nd the value of 2 2 2 2A( ) A cα β αβ α β

Q47. Find the solution of the equation x 2 4 x 6 x


www.plancess.com

Q48. Given that α, β are the roots of ax2 + bx + c = 0 and ,1 1α βα β are the of roots px2 + qx + r = 0,

�nd the relation among p, q and r in terms of a, b and c.

Q49. One side of a square is increased by x % while next side is reduced by x % to form a rectangle. �e area of the rectangle is 4 % less than the area of the original square. Find x.

Q50. On a windy day a boy rides to a place 24 km away and returns by the same road. �e wind adds 2 km per hour to his speed on his outward journey and retards him by the same amount on his way home. He takes one hour more for his return journey than for the onward journey. At what rate does he ride when there is no wind?

Q51. B takes 16 days less than A to do a piece of work. If both working together can .do it in 15 days, in how many days will B alone complete the work?

Q52. Solve the equation: 22

1 16 x 25 x 12 0

xx


www.plancess.com

EXERCISE 2 – For Competitive Examinations

Multiple Choice QuestionsDirections: �is section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q1. One of the two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2: �e other copied the constant term and coe�cient of x2

correctly as -6 and 1 respectively. �e corrects roots are

(a) 3, -2 (b) -3, 2

(c) -6, -1 (d) 6, -1

Q2. What is the condition for one root of the quadratic equation 2ax bx c 0 to be twice the other?

(a) 2b 4ac (b) 22b 9ac

(c) 2 2c 4a b (d) 2 2c 9a b

Q3. If ,α β are the roots of the equation 2ax bx c 0 , then ?a b a bα ββ α

(a) 2/a (b) 2/b

(c) 2/c (d) -2/a

Q4. If , ,α β are the roots of the equation 3 22x 3x 6x 1 0 , then 2 2 2α β is equal to

(a) -15/4 (b) 15/4

(c) 9/4 (d) 4

Q5. If 2 2x y 25 , x y = 12, then x=

(a) {3, 4} (b) {3, -3}

(c) {3, 4, -3, -4} (d) {-3, -3}

Q6. If x 7 4 3 , then 1x

x(a) 4 (b) 6

(c) 3 (d) 2

Q7. If the roots of the given equation 22x 3( 2)x 4 0λ λ are equal in magnitude but opposite in sign, then λ =

(a) 1 (b) 2

(c) 3 (d) 2/3


www.plancess.com

Q8. If the roots of the equation 2px 2qx r 0 and 2qx 2 pr x q 0 be real, then

(a) p = q (b) 2q pr

(c) 2p pr (d) 2r pq

Q9. �e equation 22x 2(p 1) x p 0 , where p is real, always has roots.

(a) Equal (b) Equal in magnitude but opposite in sign

(c) Irrational (d) Real

Q10. If the ratio of the roots of the equation 2x qx r 0 , then

(a) 2 2r b qc (b) 2 2r c qb

(c) 2 2c r q b (d) 2 2b r q c

Q11. If a b c 0 and a, b, c are rational, then the roots of the equation:2(b c a)x (c a b)x (a b c) 0

(a) Rational (b) irrational

(c) Imaginary (d) equal

Q12. If one root of the quadratic equation 2ax bx c 0 is the reciprocal of the other, then

(a) b = c (b) a = b

(c) ac = 1 (d) a = c

Q13. �e roots of the equation 1 1x 3 ,x 0

x 3(a) 3, 1 (b) 1

3,3

(c) 13,

3 (d) 1

3,3

Q14. If the equation 22x 6x p 0 has real and di�erent roots, then the values of p are given by

(a) 9p

2 (b) 9

p2

(c) 9p

2 (d) 9

p2

Q15. �e roots of 2x bx c 0 are each decreased by 2. �e resulting equation is 2x 2x 1 0 . �en

(a) b = 6 , c = 9 (b) b = 3 , c = 5

(c) b = 2 , c = -1 (d) b = -4 , c = 3

Q16. If one root of 2ax bx c 0 is equal to nth power of the other then n 1/(n 1) n 1/(n 1)(ac ) (a c)

(a) 1 (b) -1

(c) b (d) –b


www.plancess.com

More than One CorrectDirections: �is section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and

(d) out of which ONLY OR MORE may be correct.

Q17. Which of the following equation have no real roots?

(a) 2x 2 3x 5 0 (b) 22x 6 2 11 0

(c) 2x 2 3x 5 0 (d) 22x 6 2x 9 0

Q18. Two number whose sum is 8 and the absolute value of whose di�erence is 10 are roots of the equation

(a) 2x 8x 9 0 (b) 2x 8x 9 0

(c) 2x 8x 9 0 (d) 2( x 8x 9) 0

Q19. �e equality 2b 5 9b 12 is satis�ed if

(a) b > 9 (b) b < 1

(c) b > 0 (d) b < 0

Q20. �e value of m so that the equation 23x 2mx 4 0 and x (x – 4 m) + 2 = 0 may have a common root is-

(a) 1/ 2 (b) 1 / 2

(c) 1 / 2 (d) 1 / 2

Passage Based QuestionsDirections: Study the given passage (s) and answer the following questions.

Passage I

Q21. Let us consider a quadratic equation 2 2x 3ax 2a 0 . If the above equation has roots ,α β and it is given that 2 2 5α β

(i) Value of a is

(a) 1 (b) -1

(c) 1 (d) none of these

(ii) Value of D for the above quadratic equation is

(a) D > 0 (b) D < 0

(c) D = 0 (d) none of these

(iii) Product of roots is

(a) 2 (b) 1

(c) -3 (d) 3


www.plancess.com

Passage II

Q22. Let us consider a quadratic equation 2x x 1.25 0λ λ where λ is a constant. �e value of λ, such that the above quadratic equation has

(i) Two distinct roots

(a) 5λ (b) 1λ

(c) 5λ or 1λ (d) None of these

(ii) Two coincident roots

(a) 5λ or 1λ (b) 1λ or 5λ

(c) 5λ or 1λ (d) None of these

Assertion and ReasonDirections: Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the questions on the basis of following options. You have to select the one that best describes the two statements.

(a) Both Assertion and Reason are correct and reason is the correct explanation of Assertion

(b) Both assertion and reason are correct, but Reason is not the correct explanation of Assertion.

(c) Assertion is correct but Reason is incorrect.

(d) Assertion is incorrect but Reason is correct.

Q23. Assertion: If roots of the equation 2x bx c 0 are two consecutive integers, then 2b 4c 1.

Reason: If a, b, c are odd integer then the roots of the equation 2 24abc x (b 4ac)x b 0 are real and distinct.

Q24. Assertion: If 1 a 2 then a 2 a 1 a 2 a 1 2

Reason: 1 a 2 then (a-1) > 1.

Q25. Assertion: If one root is 3 2 , then the equation of lowest degree with rational coe�cients 4 2x 10x 1 0 .

Reason: For a polynomial equation with rational coe�cient irrational roots occurs in pairs.

Q26. Let a, b, c, p, q be real numbers. Suppose ,α β are the roots of the equation 2x 2px q 0 and 1

,αβ

are the roots of the equation 2ax 2bx c 0 , where 2 ( 1 , 0 , 1)β

Assertion: 2 2(q q) (b ac) 0

Reason: b paor c qa


www.plancess.com

Multiple Matching QuestionsDirections: Following question has four statements (A, B, C and D) given in Column I and four statements (p, q, r and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in Column I with entries in Column II.

Q27. Match the columns

Column I Column I

(A) If ,α β are roots of 2ax bx c 0 , then one of the roots of the equation 2ax bx(x 1)

2c(x 1) 0

(p) a < 0 , b > 0

(B) If the roots of 2ax b 0 are real, then (q) Real and equal

(C) Roots of 24x 4x 1 0 (r)1ββ

(D) Roots of (x-a) (x-b) + (x-b) (x-c) +(x-c) (x-a) = 0 are always

(s) a > 0, b < 0

(t) Real

(u)1αα

Q28. D be the discriminant of the quadratic equation 2ax bx c 0

Column I Column I

(A) (p) a < 0

(B) (q) a > 0


www.plancess.com

Column I Column I

(C) (r) D < 0

(D) (s) D > 0

(t) D= 0

Subjective QuestionsDirections: Answer the following questions.

Q29. A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the di�erences of its distances from two diametrically opposite �xed gates Q and R on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Q30. If the ratio of the roots of the equation 2x 2ax b 0 is equal to that of the roots 2x 2cx d 0

then prove that 2

2

a bdc

.

Q31. A scenery ₹ R1 .A shopkeeper gives a discount of x% and reduces its price to R2. He gives a further discount of x % on the reduced price R2 to reduce it further to R3, which reduces it by ₹ 415. A customer bargains with him and takes an x % discount on R3 and buys the scenery for ₹ 3362.8. Find the original price R1 of the scenery.

Q32. A businessman bought some items for ₹ 600, keeping 10 items for himself, then sold the remaining items at a pro�t of ₹ 5 per item. From the amount received in this deal he could buy 15 more items. Find the original price of each item.

Q33. A swimming pool is �lled by three pipes with uniform �ow. �e �rst two pipes operating simultaneously, �ll the pool in the same time during which the pool is �lled by the third pipe alone. �e second pipe �lls the pool �ve hours faster than the �rst pipe and four hours slower than the third pipe. Find the time required by each pipe to �ll the pool separately.

Q34. Find the value of p if 2 2 13

4α β αβ where α and β are roots of 2x px 1 0


www.plancess.com

SOLUTIONSExercise 1 – For School Examinations

Fill in the Blanks

1. 0 2. 13, 14 3. 6 km/hr. 4. 7 and -9

5. 2x 8x 15 0 6. 2

2

b 2ac

c 7.

2 1αα

8. 2 9. 1(b q)

2 10. Same degree 11. Two

True / False12. False 13. False 14. True 15. False

16. False 17. True 18. True

Match the Following Columns 19. (A) – s, (B) – p , (C) – r, (D) – q

(A) Let the smaller number be x.

�en, 218 x x 208

2x 18x 208 0

2x 26x 8x 208 0

x(x 26) 8(x 26) 0

x 8, 262(larger number) 18 (8) 144

Larger number = 12

(B) Let the son’s present age be x.

�en, father’s age = 2x

Now, 2x 1 8(x 1)2x 8x 7 02x 7x x 7 0

(5 7)(5 1) 0

5 1,72Father's age =(7) 49


www.plancess.com

(C) Let the son’s age be x

Father’s age = 2x2x 5x 662x 5x 66 02x 11x 6x 66 0

x(x 11) 6(x 11) 0

x 6,11 2Father's age = (6) 36

Very Short Answers Questions

20. 2x 2x 15 0

21. x 0

22. 22x 2x 1 0

23. 2px 3x 4 0 a p, b 3, c 4⇒2D b 4ac 9 4p ( 4) 9 16 p

9 16p 0 ( for real roots, D 0 )⇒

916 p 0 9 p

16⇒ ⇒

24. Domain : x R

Given inequality is equivalent to

2

2

3x 7x 82 0

x 12 2

2

3x 7x 8 2x 20

x 12

2 2

(x 1) (x 6)x 7x 60 0

x 1 x 1⇒ x [1, 6]⇒

25. x 1 or x 3 [ (3x 1) 5 x, etc.]

26. 2 6 , 2 6

27. Let the middle number be x, then the other two numbers are x – 1 and x + 1. According to the given condition, 2 2 2x [(x 1) (x 1) ] 60


www.plancess.com

Short Answer Questions

28. Let x

x 6 6 6 .....

�en, x 6 x

2 2x 6 x x x 6 0⇒ ⇒

(x 3) (x 2) 0 x 3⇒ ⇒ Or x 2

29. Given: 1 1 1 1x 5 x 4 x 2 x 7

1 1 1 1x 5 x 2 x 7 x 4

⇒

x 2 x 5 x 4 x 7(x 5)(x 2) (x 7)(x 4)

(x 7) (x 4) (x 5) (x 2)⇒

2 2x 11x 28 x 7x 10⇒ 94x 18 x

2⇒ ⇒

30. Present ages (in years)

Father = x, then Son = 45 – x … (given)

5 years ago, Father = x - 5,

Son = 45 - x – 5 =40 - x

According to the given conditions

(x - 5) (40 - x) = 124

Son’s age = 9 year and father’s age 36 year.

31. In the �gure AB = 2 m

Let CB = x m. �en AC = (2 - x) m

Now, it is given that 2AC AB .CB

2(2 x) 2x 24 x 4x 2x

2x 6x 4 0

6 36 4 1 4 6 36 16 6 2 5x

2 2 2 3 5

But 3 5 is not possible as it is more than the total length and shows external division.

Hence, CB 3 5 m.


www.plancess.com

32. When mistake was committed by �rst one in writing constant term, he got roots as 5 and 9. But co-e�cient of x was written correctly so, sum of roots = 5 + 9 = 14. �e other persons did mistake in writing co-e�cient of x, and got roots as 12 and 4. But constant term was written correctly so, product of roots 12 4 48 . So, the correct quadratic equation is where sum of roots = 14 and product of roots = 48.

�e equation is 2x (sum of roots) x product of roots = 0

i.e. 2x 14x 48 0

33. �e given equation is:

(x 1) (x 2) (x 3) (x 4) 8 0

In such type of equation we combine the factors in such a way that the product of two factors together gives some common polynomial. Rewriting the equation, we have

(x 1) (x 4) (x 2) (x 3) 8 0

Or 2 2(x 5x 4) (x 5x 6) 8 0

Let 2x 5x y

(y 4) (y 6) 8 0 or 2y 10y 24 8 0

y 8 or 2

But 2y x 5x

2x 5x 8 Or 2x 5x 2

Or 2x 5x 8 0 Or 2x 5x 2 0

5 25 32x

2

Discriminant < 0

So, there is no real solution.

5 25 8x

25 17

2

�ere are only two real solutions,

5 17 5 17x or x

2 2

34. p 4 or p 4

35. Hint: Discriminant 20 (b c) 4 (a b) (c a) 0⇒2 2 24a b c 4ab 2bc 4ca 0⇒

2( 2a b c) 0 2a b c 0.⇒ ⇒


www.plancess.com

36. ( a, b)

Hint: 1 1 1 1 1 1 1 1a b x a b x a b x x a b

⇒

x (a b x) b a (a b) a b(a b x)x ab (a b x)x ab

⇒ ⇒

21 1x (a b) x ab 0

(a b x)x ab⇒ ⇒ (x a) (x b) 0⇒

37. 15 hours.

Hint: �e tank is �lled by two pipes together in 39

8 hours i.e. in 75

8 hours

�e part of tank �lled in one hour 875

.

Let the time taken by the pipe of larger diameter to �ll the tank separately be x hours, then21 1 8

4x 35 x 375 0x x 10 75

⇒

24x 60x 25x 375 04x(x 15) 25(x 15) 0(x 15)(4 x 25) 0

25x 15 or x

4Time can not be ve.

25x is not acceptable.

4

38. If 2ax bx c 0 and 2a'x b'x c' 0 have a common root,

�en it is ca' ac'ab' ba'

Here, a = 1, b = 1 , c = -k and a’ = 1

b ‘ = - 10, c’ = (2 k - 3)

So, ( k)(1) 1(2k 3)3

1( 10) 1(1)

3k 3k 2k 33 3

10 1 11⇒ ⇒ 3k 3 33 k 12⇒ ⇒

39. -3, -15, 3, 15

40. Hint: (i) Let the two digit number be 10x+y

(ii) Given x y = 18 and


www.plancess.com

10x y 27 10y x y x 3⇒

(iii) Substitute y x 3 in �rst equation and solve for x and y.

Long Answer Questions41. Let the total number of girls who planned the picnic be = x

Total budget for eatables = ₹ 2400

Original share of money for every girl = ₹ 2400x

�e actual number of girls who attend the picnic = (x – 10)

New share of money for the girls attending the picnic =₹ 2400x 10

�e di�erence between two shares of money

= ₹ 8

2400 2400 2400x 2400(x 10)8 8

x 10 x x(x 10)⇒

224000 8x(x 10) 3000 x 10x⇒ ⇒

2

2

x 10x 3000 0

x 60x 50x 3000 0

⇒

⇒x(x 60) 50(x 60) 0⇒

(x 60) (x 50) 0⇒ x 60 , 50

Number of girls = 60

�e number of girls cannot be negative

42. Let the usual speed of the plane by x km/hour

Increased speed= (x 400) km/h

Distance DistanceSpeed= Time

Time speed⇒

Time taken by the plane to cover 1600 km. with usual speed 1600x

hours.

Again time taken by the plane to cover 1600 km. with increased speed= 1600hours

x 400

According to given information, we get,

1600 1600 40 x 400 x 21600

x x 400 60 x(x 400) 3⇒

2x 400x 960000 0⇒


www.plancess.com

(x 1200)(x 800) 0⇒

x 800 (x 1200 not permissible)

Hence, usual speed of plane = 800 km/ hr.

43. Let the time taken by B to do the work = x days.

Given that A takes 12 days less than B to do work.

Time taken by A to do the work (x 12) days

Given also that the time taken by A and B = 8 days.

Work done by A in 1 day 1x 12

Work done by B in 1 day = 1/ x

And work done by A and B in 1 day = 1 / 8

21 1 1 x 12 x 1x 28x 96 0

x x 12 8 x(x 12) 8⇒ ⇒ (x 24)(x 4) 0 x 24⇒ ⇒ or x 4

Since time taken by A = (x 12) days, x 12

x 24

Time taken by B = 24 days.

44. Here, ba

α β and ca

αβ

If roots are 1 1,α β

β α, then sum of roots are

1 1 b( ) (a c)

acα β

α β α ββ α αβ

And product = 1 1α β

β α1 c a

1 1 2a c

αβαβ

2 2 22ac c a (a c)ac ac

Hence, required equations given by2

2 a (a c)x (a c)x 0

ac ac

2 2acx (a c)bx (a c) 0⇒

45. We have p q x q r x r p x 4xr p q p q r

p q r x p q r x p q r x 3x3

r p q p q r

1 1 1 p q r x(p q r x) 3

p q r p q r⇒


www.plancess.com

1 1 1 3(p q r x) 0

p q r p q r⇒ x p q r⇒

46. Since α and β are roots of the equation

2 2 2 2A(x m ) Amx cm x 0 2 2 2or (A cm )x Amx Am 0 ......(1)

2

Am

A cmα β and

2

2

Am

A cmαβ

Now, 2 2 2 2A( ) A cα β αβ α β 2 2 2A[( ) 2 ] A cα β αβ αβ α β

2 2 2 2 2 2 4

2 2 2 2 2 2

A m 2Am A m cA mA

(A cm ) A cm A cm (A cm )

3 2 2 2 2 2 2 2 2 4

2 2

A m 2A m (A cm ) A m (A cm ) cA m

(A cm ) 2 2

00

(A cm )

47. �e equation is de�ned for x 2 0 ,4 x 0

And 6 x 0

x 2, x 4⇒ And x 6

2 x 4

Now, the given equation is x 2 4 x 6 x

Squaring both sides, we obtain

x 2 4 x 2 (x 2)(4 x) 6 x

2 (x 2)(4 x) (4 x)⇒

2 x 2 4 x

Again, squaring both sides, we get,

⇒

⇒ ⇒

4(x 2) 4 x 4 x 8 4 x12

5x 12 x5

But 2 x 4

Solution of the original equation is 12x

5

48. b c,

a aα β αβ

q1 1 pα βα β

And r1 1 pα βα β


www.plancess.com

(1 ) (1 ) q(1 )(1 ) p

α β β α⇒

α β and

r1 p

αβα β αβ

2 q1 pα β αβ

⇒α β αβ

and r1 ( ) p

αβα β αβ

b 2cqa a

pb c1

a a

⇒ and c

rapb c

1a a

b 2c qa b c p

⇒ and c ra b c p

p(b 2c) pca b c

q r⇒

p(b 2c) pcr(b 2c) qc

q r⇒ ⇒

49. Let the side of the square = a units

New length axa

100,

new breadth axa

100

Area of rectangle= 2 2

2ax ax x aa a a

100 100 10000

Area of square = 2a . Given condition is2 2

2 2 2x a 4a a a

10000 100

2 2 2 22 10000a x a 4a

a10000 100

⇒ 2 2 2 2 210000a 10000a x a 4a

10000 100⇒

2 2 2 2x a 4a x 410000 100 100 10

⇒ ⇒ 2x 400 x 20⇒ ⇒

50. Let x km per hour be the rate at which he rides when there is no wind.When there is wind, rate at which he rides on the outward journey = (x + 2) km per hourRate at which he rides in the return journey = (x- 2) km/ hr

Time taken for the outward journey 24x 2

hr Time taken for the return journey 24x 2

hours24 24

1x 2 x 2


www.plancess.com

Multiplying both sides by (x 2)(x 2)

24(x 2) 24(x 2) (x 2)(x 2)

i.e., 2 296 x 4 x 100⇒

x 10

the boy rides at the rate of 10 km per hour when there is no wind.

51. 24 days

Hint: Let A complete the work in x days, then will B complete it is (x – 16) days. According to given condition, 1 1 1

x x 16 15

52. 22

1 16 x 25 x 12 0

xx

21 1

6 x 2 25 x 12 0x x

⇒

Substituting 1x y,

x we get,

2 26(y 2) 25(y) 12 0 6y 25y 24 0⇒

26y 16 y 9y 24 0⇒

8 3 1 8(3y 8)(2y 3) 0 y , x

3 2 x 3⇒ ⇒ ⇒

23x 8x 3 0 (x 3)(3x 1) 0⇒ ⇒

1x 3,

3⇒ or

21 3 x 1 3x ,

x 2 x 2⇒

2 22x 2 3x 2x 3x 2 0 (x 2)(2x 1) 0⇒ ⇒ ⇒1

x 2,2

⇒ . So roots of the equation are,

1 12, , 3 ,

2 3

Exercise 2 – For Competitive Examinations

Multiple Choice Questions

1. (d) Let ,α β be the roots of the equation. �en 5α β and 6αβ So, the equation is 2x 5x 6 0 . �e roots of the equation are 6 and – 1.

2. (b) b2

aα α and c

2a

α α


www.plancess.com

b b3

a 3a⇒ α ⇒ α

and 2

2 c b c2 2

a 3a aα ⇒

22

2

2b c2 b 9ac

a9a⇒ ⇒

Hence, the required condition is 22b 9ac

3. (d) b c,

a aα β αβ and

22 2

2

(b 2ac)

aα β

Now (a b) (a b)a b a b (a b)(a b)

α α β βα ββ α β α

2

2 2 2

2 22 2

(b 2ac) ba b

aa( ) b( ) ac ba ab( ) b a ab ba a

α β α β

αβ α β

2 2

2 2 2 2

b 2ac b 2ac 2aa c ab ab a c

4. (a) Given equation 3 22x 3x 6x 1 0

3 1, , 3

2 2α β αβ αβ

2 2 2 2( ) ( ) 2 ( )α β α β αβ2

3 9 152.3 6

2 4 4

5. (c) 2 2x y 25, xy 12

22 4 212

x 25 x 144 25 x 0x

⇒ ⇒

2 2 2(x 16) (x 9) x 16⇒ ⇒ and 2x 9

x 4⇒ and x 3

6. (a) We have, x 7 4 3

1 1 7 4 37 4 3

x 7 4 3 7 4 3. 7 4 3


www.plancess.com

2 22 2

2 2

1x 7 4 3 7 4 3

x

4 3 4 3 4 3 4 3

2 3 2 2 3 2 3 2 2 3

2 3 2 3

( 3 2) (2 3) 4

Trick:2 2

27 4 3 2 3 2.2. 3 2 3 2 3

7. (b) Let roots are α and α , then sum of the roots

3( 2) 3( ) 0 ( 2) 2

2 2λ

α α ⇒ λ ⇒ λ

8. (b) Equation 2px 2qx r 0 and 2qx 2 pr x q 0 have real roots then from �rst

2 24q 4pr 0 q pr⇒ …..(1)

and from second 24 (pr) 4q 0 (for real root)2pr q⇒ …..(2)

From (1) and (2), we get result 2q pr

9. (d) �e discriminant of a quadratic equation2ax bx c 0 Is given by 2b 4ac .

a 2, b 2(p 1) And c = p2 2[2(p 1)] 4(2p) 4 (p 1) 8p⇒

2 24 [(p 1) 2p] 4[(p 2p 1) 2p]⇒ ⇒ 24(p 1)⇒

For any real value of p, 2(p 1) will always be positive as 2p cannot be negative for real p.

Hence, the discriminant 2b 4ac will always be positive. When the discriminant is greater than ‘0’ ir is positive, then the roots of a quadratic equation will be real.

10. (d) Let 1, 2 be the roots of equation (1) and 2, 4 be the roots of equation (2).

Equations are 2x 3x 2 0 and 2x 6x 8 0 .

Comparing with 2x bx c 0 and 2x qx r 0

We get b = - 3, c = 2 , q = - 6 and r = 8.

Putting these values in the options, we �nd that option (d) is satis�ed.


www.plancess.com

11. (a) We have, 2D (c a b) 4(b c a) (a b c)

2(a b c 2b) 4(a b c 2a) (a b c 2c) 2 2( 2b) 4( 2a)( 2c) 4(b 4ac)2 24[( a c) 4ac] 4(a c) 2{2(a c)} perfect square

12. (d) Hint: If one root is α , then the other is 1α

1αα

= product of roots c c1 a c

a a⇒ ⇒

13. (b) Hint: Check that 3 and 13

both satisfy the given equation.

14. (a) Hint: 2 2'b 4ac' ( 6) 4.2p 36 8p⇒ ⇒

98p 36 p

2⇒ ⇒ .

15. (a) Hint: b, cα β αβ

( 4) b 4⇒ α β

( 2) ( 2) 2( ) 4α β αβ α β c 2b 4

Now, 2 b 4; 1 c 2b 4

16. (d)

More than One Correct17. (a, b)

18. (b, d)

Let the one root be α then other one is 8 α .

| (8 ) | 10α α2(2 8) 100α

2

2

2 2

4 32 36 0

8 9 0

i.e. x 8x 9 0 or ( x 8x 9) 0

α α

α α

19. (a, d) Given equality is satis�ed if b 9 or b 0

20. (a, b) Let α be the common root

�en 23 2m 4 0α α and 2 4m 2 0α α

By cross-multiplication, we get2 1

4m 16m 4 6 12m 2mα α


www.plancess.com

2 120m 10 10mα α

⇒

2 12m 1 mα α

⇒

Or 22m 1 1m

2

Passage Based Questions21. Passage I

(i). (c) 3aα β22aαβ

2 2 5α β2( ) 2 5α β αβ

2 29a 2(2a ) 525a 5

a 1

(ii) (a) 2 2 2 2 2(3a) 4(2a ) 9a 8a a 1 0

(iii) (a) 22a 2(1) 2αβ

22. Passage II

�e given equation is2x x 1.25 0λ λ

a 1, b , c 1.25λ λ2 2b 4ac 4 1.( 1.25)λ λ

2 4 5 ( 5)( 1)λ λ λ λ

(i). (c) �e equation has two distinct roots if2b 4ac 0

( 5)( 1) 0λ λ

Either -5 >0⇒ λ and 1 0λ

5⇒ λ and 1λ

5⇒ λ

5 0⇒ λ and 1 0λ

5λ and 1λ

1⇒ λ


www.plancess.com

Hence the given equation has two distinct roots for

5λ or 1λ

(ii). (a) �e equation has two coincident roots if2b 4ac 0 ,

( 5) ( 1) 0λ λ

Either 5 0, 5⇒ λ λ

1 0 1⇒ λ ⇒ λ

5λ or 1

Hence the given equation has coincident roots for

5λ or 1

Assertion and Reason23. (a) Assertion: Given equation 2x bx c 0

Let ,α β be two roots such that | | 1α β2( ) 4 1⇒ α β αβ

2b 4c 1⇒

Reason: Given equation:2 24abc x (b 4ac)x b 0

2 2 2D (b 4ac) 16ab c2 2D (b 4ac) 0

Hence roots are real and unequal.

24. (c) If 1 a 2 0 a 1 1⇒

a 2 a 1 a 2 a 1⇒

1 a 1 1 a 1 2

Statement 1 is true. Statement 2 is false.

25. (a) 2 2 2x 3 2 ;x 5 2 6 ;(x 5) 24

4 2 4 2x 10x 25 24 x 10x 1 0⇒

For polynomial equation with rational coe�cients, irrational roots occur in pairs.

26. (b)


www.plancess.com

Multiple Matching Questions27. (A) –r, u ; (B) – p, s ; (C) –q; (D) – t

28. (A) – q, s ; (B)-p, s ; (C) – q, r ; (D) – p ,t

Subjective Questions29. We �rst draw the diagram, as below.

Let P be the required location of the pole. Let the distance of the pole from the gates R x m, i.e.,, RP = x m. Now the di�erence of the distances of the pole from two gates = QP – RP (or, RP - QP) = 7 m. �erefore, QP = (x + 7) m.

Now, QR= 13 m, and since QR is a diameter,QPR 90

�erefore, 2 2 2QP PR QR (By Pythagoras theorem)

i.e. 2 2 2(x 7) x 13

i.e. 2 2x 14x 49 x 169

i.e. 22x 14x 120 0

So, the distance ‘x’ of the pole from gate R satis�es the equation2x 7x 60 0

Discriminant, 2D 7 4 60 289 0

So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.

Solving the quadratic equation 2x 7x 60 0 s, by the quadratic formula, we get 7 289 7 17

x2 2

�erefore, x = 5 or -12

Since x is the distance between the pole and the gate R, it must be positive

�erefore, x = -12 is ignored. So, x = 5. So, RP = 5 and QP = 12.

Taking RP – QP = 7 or QP = x – 7

We get x = 12, -5 and take x = 12 ⇒QP = 12 i.e. QP = 12 and RP = 5

�us, the pole has to be erected on the boundary of the park at a distance of 5 m from gate R and 12 m from the gate Q or 12 m from gate R and 5 m from gate Q.

30. Consider the equation 2x 2ax b 0 ….(i)

Roots of this equation are 2a a b and 2a a b

Similarly, roots of equation 2x 2cx d 0


www.plancess.com

Are, 2c c d and

2c c d ….(ii)

Now, ratio of roots of �rst equation = ratio of roots of second equation.2 2

2 2

a a b c c d

a a b c c d⇒

Using componendo and dividend, we get

2 2 2 2

2 2 2 2

a a b a a b c c d c c d

a a b a a b c c d c c d

2 2 2 2

2a 2c a c

2 a b 2 c d a b c d⇒ ⇒

Squaring both the sides, we get 2 2

2 2

a c

a b c d

2 2a d c b.⇒ (By cross multiplication)

2

2

a b.

dc⇒

31. Cost of scenery = 1R

2 1x

R R 1100

A further discount of x % of 2R reduces it by ₹ 415.

2 1x x x

.R 415 .R 1 415100 100 100

⇒ ⇒ ….(i)

Further 2

3 1 1x x x

R R 1 1 R 1100 100 100

Again 2

4 1x x

R 3362.8 R 1 1100 100

3

1x

3362.8 R 1100

⇒ ….(ii)

Dividing (ii) by (i),3 2

1

1

x xR 1 1

100 1003362.88.1

xR x415 x1

100100 100

⇒

2y8.1

1 y⇒ x

using, y 1100


www.plancess.com

2y 8.1y 8.1 0⇒

28.1 8.1 4 1 ( 8.1) 8.1 9.9y 0.9

2 2⇒

or - 9

x1 0.9

100⇒ or x

1 9100

x 0.1 100 10⇒ or x 10 100 1000

But x = 1000 is not possible.

Using, x = 10 in (i), we get 110 x

R 1 415100 100

1415 10 10

R9

⇒ ₹ 4611.1 ₹ 4611

32. Let the original price of each item = ₹ x.

�e number of items bought by the businessman 600.

x

He sells 60010

x items at the rate if ₹ (x + 5) per item.

�e total amount received by him in this deal 60010 (x 5)

x

Now, the amount required to buy 15 more item, i.e.,

60015

x items at the rate of ₹ x per item 600

x 15x

�en according to the question

600 600x 15 10 (x 5)

x x

3000600 15x 600 10x 50

x⇒

300025x 50 0

x⇒ 225x 50x 3000 0⇒

2x 2x 120 0⇒ (x 12) (x 10) 0⇒

x 12 0⇒ Or x 10 0

x 12 Or x 10 .

But the price of an item cannot be –ve.

Hence the original price of each item is ₹ 10.


www.plancess.com

33. Let the time taken by the second pipe alone to �ll the pool be x hrs.

As the second pipe �lls the pool 5 hours faster than the �rst pipe so time taken by the �rst pipe to �ll the pool is the same as that taken by third pipe alone.

1 1 1x 5 x x 4

x x 5 1x(x 5) x 4

⇒

(x 4)(2x 5) x(x 5)⇒2 22x 5x 8x 20 x 5x⇒

2x 8x 20 0⇒2x 10x 2x 20 0⇒

x(x 10) 2(x 10) 0⇒

(x 10)(x 2) 0⇒

x 10, 2⇒

But x 2 (time can not be -ve)

x 10

Hence, time taken by �rst pipe = x + 5 = 15 hrs.

Time taken by second pipe = x = 10 =hrs.

Time taken by third pipe = x – 4 = 6 hrs.

34. Given 2 2 1 133

4 4α β αβ ….(i)

If α and β are roots of the equation 2x px 1 0

Sum of roots pα β

Product of roots, 1α β

Substituting 1α β in (1), we get 2 2 131

4α β

2 2 13 171

4 4⇒ α β

2 172

4⇒ α β αβ i.e. 2 17

24

α β

1αβ

Or 2 17 25( ) 2

4 4α β

5p

2α β

5p

2⇒

3. QUADRATIC EQUATION - TopperLearning

Documents

Transcript of 3. QUADRATIC EQUATION - TopperLearning