QUADRATIC EQUATIONS 4 - Books+App

CBSE Mathematics Question Bank Class 10

-5-TEACHERS FORUM

QUADRATIC EQUATIONS4EXERCISE 4.1, 4.2

Topic : Solution of a quadratic equation by factorisation method.

2 MARKS

1. Solve for x: 6x2 + 11x + 3 = 0 (2020-S)

Ans. 6x2 + 11x + 3 = 0

6x2 + 9x + 2x + 3 = 0

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

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

x = -3/2, x = -1/3

2. Find the roots of the quadratic equation : 3x2 - 4√3 x + 4 = 0. (2020-B)

Ans. 3x2 - 2√3 x - 2 √3 x + 4 = 0

√3 x (√3 x - 2) - 2 (√3 x - 2) = 0

(√3 x - 2) (√3 x - 2) = 0

⇒ x = 2/√3

3. Solve for x : x + 3x + 2 =

3 x - 72 x - 3, x ≠ -2,

32

Ans. (x + 3)(2x – 3) = (3x – 7)(x + 2) (2017)

⇒ 2x2 - 3x + 6x - 9 = 3x2 + 6x - 7x - 14

⇒ x2 – 4x – 5 = 0

⇒ x2 - 5x + 1x - 5 = 0

⇒ x (x - 5) + 1 (x - 5) = 0

⇒ (x – 5) (x + 1) = 0

⇒ x = 5, – 1

4. A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.

Ans. Let the ten’s digit be x and unit’s digit be y (2016)

∴ The number = 10x + y

According to the problem,

10x + y = 4 (x + y)

6x = 3y

y = 2x →(1)

Again 10x + y = 3xy

i.e., 10x + 2x = 3x (2x) [ From (1)]

12x = 6x2

⇒ 6x = 12

x = 2

(1) ⇒ y = 4

∴ The required number is 24

-6-TEACHERS FORUM

CBSE Mathematics Question Bank Class 105. Solve the following quadratic equation

for x: 4x2 - 4a2x + (a4 - b4) = 0 (2015)

Ans. The given quadratic equation can be written as

4x2 - 4a2x + (a4 - b4) = 0

i.e., (4x2 - 4a2x + a4) - b4 = 0

i.e., (2x - a2)2 - (b2)2 = 0

i.e., (2x - a2 + b2)(2x - a2 - b2) = 0

i.e., 2x = a2 - b2, or 2x = a2 + b2

⇒ x = a2 - b2

2 , x = a

2 + b2

2

6. If -5 is a root of the quadratic equation 2x2 + px - 15 = 0, find the value of p.

Ans. ƒ(x) = 2x2 + px - 15 = 0 (2014)

Since -5 is a root,

2(-5)2 + p(-5) - 15 = 0

2(25) - 5p - 15 = 0

35 - 5p = 0

⇒ p = 35/5 = 7

7. Find the two consecutive odd positive integers sum of whose squares is 290.

Ans. Let the two consecutive odd integers be x and x + 2. (2012)

Given, x2 + (x + 2)2 = 290 x2 + x2 + 4x + 4 = 290 2x2 + 4x - 286 = 0

i.e., x2 + 2x - 143 = 0 13 x -11 = -143

x2 + 13x - 11x - 143 = 0 13 + -11 = 2

x (x + 13) - 11 (x + 13) = 0 (x + 13) (x - 11) = 0 i.e., x = -13 or 11

Since x is a positive integer, x cannot be -13.

So x = 11 and x + 2 = 13

The two odd positive integers are 11 and 13.

3 MARKS

8. Solve for x: 1 x + 4

- 1 x - 7

= 11 30

, x ≠ -4, 7

Ans. 1 x + 4

- 1 x - 7

= 11 30

(2020 - S, 2019)

x - 7 - x - 4 (x + 4) ( x - 7)

= 11 30

⇒ -1x2 - 3x - 28

= 1 30

⇒ x2 - 3x + 2 = 0

⇒ (x - 2) (x - 1) = 0

∴ x = 2 or 1

9. The product of two consecutive positive integers is 306.

Find the integers. (2020-B)Ans. Let two consecutive integers x, x + 1

x (x + 1) = 306 ⇒ x2 + x - 306 = 0

x2 + 18x - 17x - 306 = 0

x (x + 18) - 17 (x + 18) = 0

⇒ (x + 18) (x - 17) = 0

⇒ x = -18 (Rejected), 17

∴ Two consecutive integers 17, 18

10. Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m2. (2019)

Ans. Let l be the length and b be the breadth of the park

∴ 2 (l + b) = 60

⇒ l + b = 30 ............... (1)

and l × b = 200

-7-TEACHERS FORUM

⇒ l (30 – l) = 200 [From (1)]

⇒ l2 – 30l + 200 = 0

⇒ (l – 20) (l – 10) = 0

⇒ l = 20 or 10

So length = 20 m, breadth = 10 m

11. Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides of the two squares.

Ans. Let x and y be length of the sides of two squares. (2019)

∴ x2 + y2 = 157 ............... (1)

and 4 (x + y) = 68 ⇒ x + y = 17

⇒ y = 17 - x .............. (2)

(1) ⇒ x2 + (17 – x)2 = 157

x2 + 289 + x2 – 34x – 157 = 0

x2 – 17x + 66 = 0

(x – 6) (x – 11) = 0

⇒ x = 6 or 11

∴ (2) ⇒ y = 11 or 6

Hence length of sides of squares are 6 m and 11 m.

12. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed. (2018)

Ans. Let the usual speed of the plane be x km/hr.

Normal time taken = distancespeed = 1500

x

When the speed is increased by 100

km / h, the time taken = 1500

x + 100


1500

x -

1500x + 100 =

3060

⇒ 1500 (1x - 1

x + 100 ) = 12

x + 100 - xx (x + 100) =

13000

⇒ 100

x2 + 100x =

13000

⇒ x2 + 100x - 300000 = 0

⇒ x2 + 600x - 500x - 300000 = 0

⇒ x (x + 600) - 500 (x + 600) = 0

⇒ (x + 600)(x - 500) = 0

⇒ x ≠ - 600, x = 500

∴speed of plane = 500 km/hr

13. Find the roots of quadratic equation √3x2 – 2√2 x – 2√3 = 0. (2017, 2015)

Ans. √3 x2 - 2√2 x - 2√3 = 0

⇒ √3 x2 - 3√2 x + √2 x - 2√3 = 0

⇒ √3 x (x - √6 ) + √2 (x - √6 ) = 0

⇒ (√3 x + √2 ) (x - √6 ) = 0

√6⇒ x = ,

√3√2-

14. Solve for x : √2 x2 + 7x + 5√2 = 0 (2017)

Ans. Given, √2 x2 + 7x + 5√2 = 0

⇒ √2 x2 + 2x + 5x + 5√2 = 0

√2 x (x + √2 ) + 5(x + √2 ) = 0

⇒ (√2 x + 5)(x + √2 ) = 0

⇒ x = -√2 or x = -5√2

Quadratic Equations


-8-TEACHERS FORUM

15. Find the roots of quadratic equation 2x2 – 5x + 3 = 0 by completing the square method. (2017)

Ans. 2x2 – 5x + 3 = 0

or x2 - 52 x +

32 = 0

⇒ (x - 54 )2

- 2516 +

32 = 0

⇒ x - ( 54 )2

= 116

⇒ (x - 54 )2

=( 14 )2

⇒ x - 54 =

14 ⇒ x =

32

or x - 54 =

-14 ⇒ x = 1

16. Solve : 1(x - 1)(x - 2)

+ 1(x - 2)(x - 3)

= 23

, x ≠ 1, 2, 3. (2016)

Ans. (x - 2)(x - 3) + (x - 1)(x - 2)(x - 1)(x - 2)(x - 2)(x - 3)

= 23

(x - 3) + (x - 1)(x - 1)(x - 2)(x - 3)

= 23

2x - 4(x - 1)(x - 2)(x - 3)

= 23

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

= 23

2(x - 1)(x - 3)

= 23

i.e., x2 - 4x + 3 = 3

x2 - 4x = 0

x (x - 4) = 0

⇒ x = 0, x = 4

17. Solve for x : (2016)

2x x - 3

+ 12x + 3

+ 3x + 9 (x - 3)(2x + 3)

= 0, x ≠ 3, -3/2

Ans. 2x (2x + 3) + 1 (x - 3) (x - 3)(2x + 3)

+ 3x + 9 (x - 3)(2x + 3)

= 0

4x2 + 6x + x - 3 + 3x + 9 (x - 3)(2x + 3)

= 0

4x2 + 10x + 6 = 0

⇒ 2x2 + 5x + 3 = 0

⇒ (x + 1)(2x + 3) = 0

⇒x = -1, x = -32

18. Find the roots of the quadratic equation: a2 b2 x2 + b2x - a2x - 1 = 0. (2012, 2013)

Ans. Given, a2 b2 x2 + b2x - a2x - 1 = 0

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

(a2x + 1) (b2x - 1) = 0

i.e., a2x + 1 = 0 or b2x - 1 = 0

i.e., a2x = -1 or b2x = 1

⇒ x = -1a2 or x = 1

b2

19. The length of a rectangular plot is greater than thrice its breadth by 2 m. The area of the plot is 120 sq. cm. Find the length and breadth of the plot.

Ans. Let breadth of the rectangle = x m

then length = 3x + 2 (2012)

Given, area = 120 cm2

i.e., (3 x + 2) x = 120


-9-TEACHERS FORUM

3 x2 + 2x - 120 = 0 20 x -18 = -360

3 x2 + 20x - 18x - 120 = 0 20 + -18 = 2

x (3x + 20) - 6 (3x + 20) = 0

(3x + 20)(x - 6) = 0

3x + 20 = 0

or x - 6 = 0

i.e., x = -203 or x = 6

breadth cannot be negative. ∴ x = 6 m

then length = 3x + 2 = 3 x 6 + 2 = 20 m

20. Solve for x : (2011, 2013)

12 abx2 - (9a2 - 8b2) x - 6ab = 0

Ans. Given, 12 abx2 - (9a2 - 8b2) x - 6ab = 0

12abx2 - 9a2x + 8b2x - 6ab = 0

3ax (4bx - 3a) + 2b (4bx -3a) = 0

(4bx - 3a)(3ax + 2b) = 0

i.e, 4bx = 3a or 3ax = -2b

x = 3a4b

or x = -2b3a

21. 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. (2012)

Ans. Let the smaller side be x cm and larger side be y cm.

According to first condition,

x2 + y2 = (3√5 )2

i.e., x2 + y2 = 45

y2 = 45 - x2 →(1)

According to the second condition,

(3x)2 + (2y)2 = 152

9x2 + 4y2 = 225 →(2)

Sub. (1) in (2)

⇒ 9x2 + 4 (45 - x2) = 225

9x2 + 180 - 4x2 = 225

⇒ 5x2 + 180 = 225

5x2 = 45

x2 = 9

x = ± 3

Since side cannot be -ve x = 3

then (1) ⇒ y2 = 45 - 9 = 36

i.e., y = 6

longer side = 6 cm and

smaller side = 3 cm

22. A two digit number is such that the product of the digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the numbers. (2011)

Ans. Let unit’s digit is x then ten’s digit = 18x

∴ Number = 180x + x

On interchanging the digits the number

becomes 10x + 18x

According to the question

180x + x - 63 = 10x + 18

x

180 + x2 - 63xx = 10 x2 + 18

x 180 + x2 - 63x = 10x2 + 18

-9x2 - 63x + 162 = 0

⇒ x2 + 7x - 18 = 0

Quadratic Equations


-10-TEACHERS FORUM

⇒ x2 + 9x - 2x - 18 = 0

(x + 9) - 2(x + 9) = 0

(x + 9) (x - 2) = 0

⇒x = -9 or x = 2 [neglect negative]

i.e., unit digit is 2, ten’s digit is 9.

∴ Number = 92

23. Find two positive numbers whose squares have the difference 48 and the sum of the numbers is 12. (2011)

Ans. Since sum of the numbers is 12, the two numbers are x and 12 - x.

According to the question,

x2 - (12 - x)2 = 48

x2 - 144 - x2 + 24x = 48

24x = 144 + 48

x = 19224

= 8

The two numbers are 8 and 4.

24. Divide 29 into two parts so that the sum of the squares of the two parts is 425.

Ans. Let the two parts be x and y. (2014)

Then, x + y = 29 ⇒ y = 29 - x


x2 + y2 = 425

x2 + (29 - x)2 = 425

x2 + 841 - 58x + x2 = 425

x2 - 29x + 208 = 0

x2 - 16x - 13x + 208 = 0

x (x - 16) - 13 (x - 16) = 0

(x - 16) (x - 13) = 0

x = 16, 13

When x = 16, y = 13

and when x = 13, y = 16

∴Two parts are 13 and 16.

25. The hypotenuse of a right angled triangle is 13 m long. If the base of the triangle is 7 m more than the other side, find the sides of the triangle.

Ans. Let one side be (x) m (2013, 2014)

∴other side = (7 + x) m

A

B C

13x

x + 7 By pythagoras theorem,

132 = x2 + (7 + x)2

169 = x2 + x2 + 49 + 14x

2x2 + 14x + 49 - 169 = 0

2x2 + 14x - 120 = 0

x2 + 7x - 60 = 0

x2 + 12x - 5x - 60 = 0

x(x + 12) - 5(x + 12) = 0

(x + 12)(x - 5) = 0

Therefore, x = -12 or 5 ,

Rejecting x = -12

One side, x = 5 m

Other side = (7 + 5) = 12 m

4 MARKS

26. The sum of the areas of two squares is 640 m2. If the difference of their perimeters be 64 m, find the sides of the two squares. (2020-S)

Ans. Let sides of larger and smaller squares


-11-TEACHERS FORUM

be x and y respectively.

x2 + y2 = 640 .......... (i)

4x - 4y = 64

⇒ x - y = 16

⇒ x = 16 + y .......... (ii)

(1) ⇒ (16 + y)2 + y2 = 640

256 + 32y + 2y2 = 640

2y2 + 32y - 384 = 0

y2 + 16y - 192 = 0

y2 + 24y - 8y - 192 = 0

y (y + 24) - 8(y + 24) = 0

(y + 24) (y - 8) = 0

⇒ y = –24 or y = 8

Thus, side of smaller square = 8 cm & that of larger square = 24 cm

27. In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The average speed of the trip was decreased by 200 km/hr and thus the time of flight increased by 30 minutes. Find the average speed of the aircraft originally. (2020-S)

Ans. Let average speed of aircraft be x km/h

A.T.Q, 600x - 200 - 600

x = 12

600 ( 1x - 200 - 1

x ) = 12

x - x + 200x2 - 200x

= 11200

x2 - 200x - 240000 = 0

x2 - 600x + 400x - 240000 = 0

x (x - 600) + 400 (x - 600) = 0

(x - 600) (x + 400) = 0

x = 600 km/h

∴ Original speed = 600 km/h

28. ` 9,000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got ` 160 less. Find the original number of persons. (2020-S)

Ans. Let original number of persons be x

A.T.Q, 9000x

- 9000x + 20 = 160

9000 ( 1x

- 1x + 20 ) = 160

x + 20 - xx2 + 20 x = 160

9000

20x2 + 20 x = 160

9000

1x2 + 20 x = 1

1125

x2 + 20x - 1125 = 0

x2 + 45x - 25x - 1125 = 0

x (x + 45) - 25 (x + 45) = 0

(x + 45) (x - 25) = 0

x = 25

∴ Number of persons = 25

29. Sum of the areas of 2 squares is 544 m2. If the difference of their perimeter is 32 m, find the sides of two squares. (2020-S)

Ans. Let ‘a’ and ‘b’ be the sides of two squares, with a > b.

Given, a2 + b2 = 54.......... (1)

and 4a - 4b = 32

or a - b = 8

∴ a = b + 8...... (2)

(1) ⇒(b + 8)2 + b2 = 544

Quadratic Equations


-12-TEACHERS FORUM

⇒ 2b2 + 16b - 480 = 0

∴ b2 + 8b - 240 = 0

⇒ b2 + 20b - 12b - 240 = 0

⇒b(b + 20) - 12(b + 20) = 0

⇒ (b + 20) (b - 12) = 0

⇒ b = 12 m

(2) ⇒ a = 12 + 8 = 20 m.

30. Three consecutive positive integers are such that the sum of the square of the first and product of the other two is 46. Find the integers. (2020-B)

Ans. Let 3 consecutive integers be

x, x + 1, x + 2

According to question,

x2 + (x + 1) (x + 2) = 46

x2 + x2 + 3x + 2 = 46

⇒ 2x2 + 3x - 44 = 0

2x2 + 11x - 8x - 44 = 0

x (2x + 11) - 4 (2x + 11) = 0

⇒ (2x + 11) (x - 4) = 0

⇒ x = 4

∴Integers are 4, 5, 6

31. Some students planned a picnic. The total budget for food was ` 2,000 but 5 students failed to attend the picnic and thus the cost for food for each member increased by ` 20. How many students attended the picnic and how much did each student pay for the food.(2020- B)

Ans. Let the no. of students planned for the picnic be ‘x’

Total amount = ` 2000

∴ cost per head = 2000x

As 5 students failed to attend, no. of students attended = x - 5

∴ cost for each student = 2000 x - 5

As per question, 2000 x - 5 - 2000

x = 20

2000 [ 1 x - 5 - 1

x ] = 20

⇒ x - x + 5x2 - 5x = 20

2000

5x2 - 5x = 1

100 x2 - 5x - 500 = 0

x2 - 25x + 20x - 500 = 0

x(x - 25) + 20(x - 25) = 0

(x + 20)(x - 25) = 0

⇒ x = -20 or x = 25

Since x cannot be negative, x = 25.

So students attended the picnic = x - 5 = 20

and cost for each student = 2000 20

= Rs.100 each

32. A train covered a certain distance at a uniform speed. If the train would have been 6 km/hr. faster, it would have taken 4 hours less than the scheduled time and if the train were slower by 6 km/hr, it would have taken 6 hrs more than the scheduled time. Find the length of the journey. (2020 - S)

Ans. Let uniform speed = x km/h

Distance travelled = d km

∴ time taken = dx

hr

Case 1:

Speed = x + 6


-13-TEACHERS FORUM

Time taken, t1 = dx + 6


dx

- dx + 6 = 4

d [1x - 1x + 6] = 4

d [ x + 6 - xx2 + 6x ] = 4

d = 4 (x2 + 6x)6 (1)

Case 2 :

Similarly, dx - 6 - d

x = 6

d [ 1x - 6 - 1

x ] = 6

d [x - x + 6x2 - 6x ] = 6

d = 6 (x2 - 6x)6

= x2 - 6x (2)

Since distance remains the same,

4 (x2 + 6x)6 = x2 - 6x

2x2 + 12x = 3x2 - 18x

x2 - 30x = 0

x (x - 30) = 0

⇒ x = 30

(2) ⇒ d = 302 - 6 x 30

= 900 - 180 = 720 km

33. The sum of the reciprocals of the ages of a child 3 years ago and 5 years

hence from now is 13 . Find his present

age. (2020 - B)Ans. Let the present age = x years

His age 3 years ago = x - 3

His age after 5 years = x + 5

Given, 1x - 3 + 1

x + 5 = 13

x + 5 + x - 3(x - 3) (x + 5) = 13

2x + 2x2 + 2x - 15 = 1

3

x2 + 2x - 15 = 6x + 6

x2 - 4x - 21 = 0

x2 - 7x + 3x - 21 = 0

x (x - 7) + 3 (x - 7) = 0

(x - 7) (x + 3) = 0

⇒ x = 7

So, Present age = 7 yrs.

34. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hr less for the same journey. Find the speed of the train. (2020 - B, 2019)

Ans. Let speed of train be x km/h

Distance travelled = 360 km

∴ time = 360x

If the speed is 5 km/h more,

time = 360x + 5


360x

- 360x + 5 = 1

360 [1x - 1x + 5] = 1

360 [ x + 5 - xx (x + 5) ] = 1

Quadratic Equations


-14-TEACHERS FORUM

360 x 5x2 + 5x

= 1

⇒ x2 + 5x – 1800 = 0

⇒ x2 + 45x - 40x - 1800 = 0

⇒ x (x + 45) - 40 (x + 45) = 0

⇒ (x + 45) (x – 40) = 0

x = -45, rejected and x = 40

So, speed of train = 40 km/h

35. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the Rectangular park and of altitude 12 m. Find the length and breadth of the park. (2020-B)

Ans. Let length of rectangle = x

∴ Breadth = x - 3

Area of reactangle

= x (x - 3) = x2 - 3x

B

E

CD

x - 3 m

x

12 m

A

Area of Isosceles

ΔADE = 12 bh

= 12 (x - 3) x 12 = 6x - 18

ATQ, Area of reactangle

= Area of Isosceles Δ+ 4

x2 - 3x = 6x - 18 + 4

x2 - 9x + 14 = 0

(x - 7) (x - 2) = 0

x = 7, x = 2 (Rejected)

∴ Length of rectangle = 7 cm

Breadth of rectangle = 4 cm

36. Two water taps together can fill a tank in

1 78 hours. The tap with longer diameter

takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. (2019)

Ans. Let the smaller tap fills the tank in x hrs

∴ the larger tap fills the tank in (x – 2) hrs.

Time taken by both the taps together

= 158 hrs.

Therefore 1x

+ 1x - 2 = 8

15

⇒ x - 2 + xx2 - 2x

= 815

⇒ 30x - 30 = 8x2 - 16x

⇒ 8x2 - 46x + 30 = 0

⇒ 4x2 – 23x + 15 = 0

⇒ 4x2 - 20x - 3x + 15 = 0

⇒ 4x (x - 5) - 3 (x - 5) = 0

⇒ (4x – 3) (x – 5) = 0

⇒ x = 34 , x = 5

x ≠ 34 ∴ x = 5

Smaller and larger taps can fill the tank seperately in 5 hrs and 3 hrs resp.

37. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the


-15-TEACHERS FORUM

difference of its distances from two diametrically opposite fixed gates A &B on the boundary is 7 m. Is it possible to do so ? If yes, at what distances from the two gates should the pole be erected?

Ans. (2019)

PB – PA = 7 m

Let AP be x m ∴ PB = (x + 7) m

AB2 = PB2 + AP2

∴ 132 = (x + 7)2 + x2

169 = x2 + 14x + 49 + x2

⇒ 2x2 + 14x - 120 = 0 ⇒ x2 + 7x – 60 = 0

⇒ (x + 12) (x – 5) = 0

∴ x = 5, –12 Rejected

∴ Situation is possible.

∴ Distance of pole from gate A = 5 m and distance of pole from gate B = 12 m.

38. Solve the following equation for x :

1x + 1

+ 2x + 2

= 7x + 5

, x ≠ -1, -2, -5

Ans. 1

x + 1 + 2

x + 2 = 7

x + 5 (2019)

⇒(x + 2) + 2(x + 1)

(x + 1)(x + 2) =

7x + 5

⇒3x + 4

x2 + 3x + 2 =

7x + 5

⇒3x2 + 4x + 15x + 20 = 7x2 + 21x + 14

⇒ 4x2 + 2x - 6 = 0

⇒ 2x2 + x - 3 = 0

⇒2x2 + 3x - 2x - 3 = 0

⇒x (2x + 3) -1 (2x + 3) = 0

⇒ (x - 1) (2x + 3) = 0

⇒ x = 1 or x = - 32

39. In a class test, the sum of Ramesh’s marks in Mathematics and English is 40. Had he got 3 more marks in Mathematics and 4 marks less in English, the product of his marks would have been 360. Find his marks in the two subjects separately. (2019)

Ans. Let marks in Mathematics be x. ∴ Marks in English = 40 – x According to the question, (x + 3) (40 – x – 4) = 360

⇒ (x + 3) (36 - x) = 360

⇒ 36x – x2 + 108 - 3x = 360

⇒ x2 – 33x + 252 = 0

⇒ (x – 21) (x – 12) = 0

⇒ x = 21 or 12

∴ Marks in Mathematics = 21 and Marks in English = 19

or Marks in Mathematics = 12 and Marks in English = 28

40. In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects. (2019)

Ans. Let marks in Hindi be x. Then marks in English = 30 – x

Given, (x + 2) (30 – x – 3) = 210

Quadratic Equations


-16-TEACHERS FORUM

(x + 2) (27 - x) = 210

⇒x2 – 25x + 156 = 0

(x – 13) (x – 12) = 0

⇒ x = 13 or x = 12

∴ Marks in English = 30 – 13 = 17

or 30 – 12 = 18

∴ Marks in Hindi & English are (13, 17) or (12, 18)

41. In a test, the sum of marks obtained by a student 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. (2019)

Ans. Let the marks obtained in Mathematics be x

∴ Marks obtained in science = 28 – x


(x + 3) (28 – x – 4) = 180

(x + 3) (24 – x) = 180

⇒ 24x - x2 + 72 - 3x = 180

⇒ x2 – 21x + 108 = 0

(x – 12) (x – 9) = 0

x = 12 or 9

If x = 12, Marks in mathematics = 12

Marks in science = 16

If x = 9, Marks in mathematics = 9

Marks in science = 19

42. Solve : 1

a + b + x = 1

a + 1

b + 1

x,

a ≠ 0, b ≠ 0, x ≠ 0 (2018)

Ans. 1 a + b + x

= 1 a

+ 1 b

+ 1x

⇒ 1 a + b + x

- 1x

= 1 a

+ 1 b

⇒ x - a - b- x

(a + b + x) x = a + b

ab

-(a + b) x(a + b + x)

= a + b ab

⇒-ab =x (a + b) + x2

⇒ x2 + (a + b)x + ab = 0

⇒ (x + a) (x + b) = 0

⇒ x = –a, – b

43. Two taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. (2019, 2018)

Ans. Let the tap of smaller diameter fills the tank in x hrs.

∴Tap of larger diameter fills the tank in (x - 9) hrs

According to the statements

1x

+ 1

x - 9 = 16

x - 9 + x x (x - 9) =

16

⇒ 6(2x - 9) = x(x - 9)

12x - 54 = x2 - 9x ⇒ x2 - 21x + 54 = 0

x2 - 18x - 3x + 54 = 0

x (x - 18) - 3 (x - 18) = 0

⇒ (x - 18) (x - 3) = 0

x ≠ 3 ∴ x = 18


-17-TEACHERS FORUM

Now the tap of larger diameter fills the tap in 18 - 9 = 9 hours

Taps can fill the tank separately in 18 hrs and 9 hrs respectively.

44. Solve for x : x + 1x - 1

- x - 1x + 1

= 56

,

x ≠ 1, -1 (2018)

Ans. (x + 1)2 - (x - 1)2

x2 - 1 = 56

x2 + 2x + 1 - (x2 - 2x + 1)

x2 - 1 = 56

6 (4x) = 5(x2 - 1)

⇒ 5x2 - 24x - 5 = 0

⇒ 5x2 - 25x + 1 x - 5 = 0

⇒ 5x (x - 5) + 1 (x - 5) = 0

⇒ (5x + 1)(x - 5) = 0

⇒ x = -15 , x = 5

45. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an av-erage speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the orig-inal average speed ? (2018)

Ans. Let the original average speed of train be x km/hr

∴Normal time = 63x

If the speed is 6 km / h more,

New time taken = 72

x + 6

As per question, 63x

+ 72

x + 6 = 3

9 (7x + 8

x + 6) = 3

7x + 42 + 8x

x2 + 6x =

13

45x + 126 = x2 + 6x

⇒ x2 - 39x - 126 = 0

⇒ x2 - 42x + 3x - 126 = 0

⇒ x (x - 42) + 3 (x - 42) = 0

⇒ (x - 42)(x + 3) = 0

⇒ x ≠ -3 ∴ x = 42

Original speed of train is 42 km/hr

46. Two pipes together can fill a tank in 12 hours. If first pipe can fill the tank 10 hours faster than the second, then how many hours will the second pipe take to fill the tank? (2018)

Ans. Let the time taken by first pipe to fill a tank = x hour

and the time taken by second pipe to fill the tank = (x + 10) hours.

As per question 1x

+ 1x + 10

= 112

(x + 10) + xx (x + 10)

= 112

⇒ (2x + 10) 12 = x (x + 10)

⇒ 24x + 120 = x2 + 10x

⇒ x2 – 14x – 120 = 0

⇒ x2 – 20x + 6x -120 = 0

⇒ x (x – 20) + 6 (x - 20) = 0

⇒ (x – 20) (x + 6) = 0

⇒ x = 20, x = -6 rejected

⇒ x = 20 hours

⇒ Second pipe can fill in 30 hours.

47. A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of slower train is 10 km/

Quadratic Equations


-18-TEACHERS FORUM

hr less than that of faster train, find the speeds of two trains. (2018)

Ans. Let the speed of faster train be x km/hr

∴ Speed of slower train = (x –10) km/hr


200x - 10

- 200x

= 1

⇒200 x - 200 x + 2000

(x - 10 ) x = 1

2000x2 -10x

= 1

⇒ x2 – 10x – 2000 = 0

⇒ x2 – 50x + 40x - 2000 = 0

⇒ x (x – 50) + 40 (x - 50) = 0

⇒ (x – 50) (x + 40) = 0

⇒x = 50, – 40 rejected

∴Speed of faster train = 50 km/ hr

Speed of slower train = 40 km/ hr

48. Solve for x :

4x

52x + 3

32

- 3 = ; x ≠ 0, -( ( (2017)

Ans. Given, 4x - 3 =

52 x + 3

⇒ 4 - 3x

x = 5

2 x + 3

⇒ (4 – 3x) (2x + 3) = 5x

⇒ 8x + 12 – 6x2 – 9x = 5x

⇒ 6x2 + 6x – 12 = 0

⇒ x2 + x – 2 = 0,

⇒ x = 1, – 2

49. Find two consecutive odd natural numbers, the sum of whose squares is

394. (2017)Ans. Let the two consecutive odd natural

numbers be x and x + 2.

Therefore x2 + (x + 2)2 = 394

⇒ x2 + x2 + 4x + 4 = 394

⇒ 2x2 + 4x – 390 = 0

⇒ x2 + 2x - 195 = 0

⇒ x2 + 15x - 13x - 195 = 0

⇒ x (x + 15) - 13 (x + 15) = 0

⇒ (x + 15) (x – 13) = 0

⇒ x = -15, 13

⇒ x ≠ -15 ∴ x = 13

Hence numbers are 13 and 15.

50. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do the work ? (2017)

Ans. Let B takes x days to finish the work, then A takes (x – 5) days to finish the work.

∴ 1x +

1x - 5 =

16

x - 5 + xx2 - 5x =

16

⇒ 6(2x – 5) = x2 – 5x

x2 – 17x + 30 = 0

⇒ x2 - 15x - 2x + 30 = 0

⇒ x (x - 15) - 2 (x - 15) = 0

⇒ (x – 15) (x – 2) = 0

∴x = 15 or x = 2.

x ≠ 2 as x > 5 ∴ x = 15

So, B can finish the work in 15 days.

51. One year ago, a man was 8 times as


-19-TEACHERS FORUM

old as his son. Now his age is equal to the square of his son’s age. Find their present ages. (2017)

Ans. Let the present age of son be x years

∴ The present age of father is x2 years

One year ago, age of son = x - 1

and age of father = x2 - 1


8(x – 1) = x2 – 1

⇒ 8x - 8 = x2 - 1

⇒ x2 – 8x + 7 = 0

x2 - 7x - 1x + 7 = 0

⇒ x (x - 7) - 1 (x - 7) = 0

⇒ (x – 7) (x – 1) = 0

⇒ x = 7, x ≠ 1

Present age of son is 7 yrs, Present age of father is 49 yrs.

52. A pole has to be erected at a point on the boundary of a circular park of diameter 17 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Find the distances from the two gates where the pole is to be erected. (2016)

Ans. Let P be the location of the pole such that its distance from gate B is x metres

∴AP = x + 7

AB is a diameter

⇒ ∠APB = 90° and AB = 17m

A

B

O

P

x

x + 7

•

∴ x2 + (x + 7)2 = 172

x2 + x2 + 14x + 49 = 289

x2 + x2 + 14x – 240 = 0

x2 + 7x – 120 = 0

x = -7 ±√49 + 4802

= -7 ±232

= 8 or -15

∴ Distance of pole from gate B = 8 m

and distance of pole from gate A = 15 m

53. Find x in terms of a, b and c :

ax - a

+ bx - b

= 2cx - c

, x ≠ a, b, c (2016)

Ans. a(x - b) + b(x - a)(x - a)(x - b)

= 2cx - c

⇒a(x - b)(x - c) + b(x - a)(x - c) = 2c(x - a)(x - b)

⇒a[x2 - xc - xb + bc] + b[x2 - xc - ax + ac]

= 2c(x2 - bx - ax + ab)

⇒ax2 - axc - axb + abc + bx2 - bcx - abx

+ abc = 2cx2 - 2bcx - 2acx + 2abc

⇒ ax2 - axc - axb + bx2 - bcx - abx - 2cx2

+ 2bcx+ 2acx = 2abc - 2abc

⇒ x2(a + b - 2c) + x (-ac - ab - bc - ab + 2bc

+ 2ac) = 0

⇒ x(a + b - 2c) + (ac + bc - 2ab) = 0

[Divide throughout by x]

∴ x = 2ab - ac - bca + b - 2c

54. Solve the following for x : (2013)

12a + b + 2x

= 12a

+ 1b

+ 12x

Quadratic Equations


-20-TEACHERS FORUM

Ans. 12a + b + 2x

- 12x

= 12a

+ 1b

⇒ 2x - 2a - b - 2x2x(2a + b + 2x)

= b + 2a2ab

⇒ - (2a + b)2x(2a + b + 2x)

= b + 2a2ab

⇒ - 1x(2a + b + 2x)

= 1ab

⇒ 2ax + bx + 2x2 = -ab

⇒ 2x2 + 2ax + bx + ab = 0

⇒ 2x (x + a) + b(x + a) = 0

⇒ (x + a) (2x + b) = 0

⇒ x + a = 0 or 2x + b = 0

⇒ x = -a or x = -b2

55. Solve : 4 (x + 1) + 4(1 - x) = 10. (2012)

Ans. Given, 4 (x + 1) + 4(1 - x) = 10

⇒ 4x . 4 + 44x = 10

⇒ 4y + 4y = 10 [Put 4x = y]

⇒ 4y2 + 4 = 10y

⇒ 4y2 - 10y + 4 = 0

2y2 - 5y + 2 = 0 -4 x -1 = 5

2y2 - 4y - y + 2 = 0 -4 + -1 = -5

2y(y - 2) -1(y - 2) = 0

⇒ (y - 2) (2y - 1) = 0

⇒ y - 2 = 0 or 2y - 1 = 0

⇒ y = 2 or y = 12

But y = 4x

Then, 4x = 2 or 4x = 12

⇒ (2)2x = 21 or (2)2x = (2)-1

⇒ 2x = 1 or 2x = -1

⇒ x = 12 or x = -1

2

56. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the length of the sides of the field. (2015)

Ans. Let the length of shorter side be x m.

A Bx

CD

x + 16

x + 14

length of diagonal = (x + 16) m

and, length of longer side = (x + 14) m

In ΔBCD, x2 + (x + 14)2 = (x + 16)2

x2 + x2 + 28x + 196 = x2 + 32x + 256

⇒ x2 - 4x - 60 = 0

⇒ x2 - 10x + 6x - 60 = 0

x(x - 10) + 6(x - 10) = 0

⇒ (x - 10)(x + 6) = 0

⇒ x = 10, x = -6

since x side cannot be negative, x = 10

∴ length of sides are 10 m and 24 m.

57. Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares. (2013)

Ans. Let the sides of the two squares be x cm


-21-TEACHERS FORUM

and y cm where x > y.

Then, their areas are x2 and y2 and their perimeters are 4x and 4y.

Given x2 + y2 = 400 →(1)

Also, 4x - 4y = 16

⇒ 4(x - y) = 16

⇒ x - y = 4

⇒ x = y + 4 →(2)

Substituting (2) in (1), we get:

(y + 4)2 + y2 = 400

⇒ y2 + 16 + 8y + y2 = 400

2y2 + 16 + 8y - 384 = 0

y2 + 4y - 192 = 0

y2 + 16y - 12y - 192 = 0

⇒ y(y + 16) - 12(y + 16) = 0

⇒ (y + 16)(y - 12) = 0

⇒ y = -16 or y = 12

Since, y cannot be negative, y = 12.

Now (2) ⇒ x = y + 4 = 12 + 4 = 16

Thus, the sides of the two squares are 16 cm and 12 cm.

58. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed If the truck covers the total distance in a 5 hours, find the first speed of the truck. (2015)

Ans. let the first average speed of truck be x km/h.

Time taken to cover 150 km, t1 = 150x

Time taken to cover 200 km, t2 = 200x + 20

According to the problem t1 + t2= 5 hr

i.e., 150x

+ 200x + 20

= 5

150x + 3000 + 200x = 5(x2 + 20x)

x2 - 50x - 600 = 0

x2 - 60x + 10x - 600 = 0

x(x - 60) + 10(x - 60) = 0

(x - 60) (x + 10) = 0

⇒ x = 60, x = -10

∴ speed = 60 km/h

59. The total cost of certain length of a piece of cloth is ` 200. If the piece was 5 m longer and each metre of cloth cost ` 2 less, the cost of the piece would have remainded unchanged. How long is the piece and what is its original rate per metre? (2015)

Ans. Let length of cloth be x m.

cost per meter = ` 200x

New length of cloth = (x + 5) m

New cost per meter = ` ( 200x

- 2) Now according to the problem,

(x + 5)( 200x

- 2) = 200

200 - 2x + 1000x

- 10 = 200

x2 + 5x - 500 = 0

⇒ x2 + 25x - 20x - 500 = 0

⇒ (x + 25) - 20(x + 25) = 0

x(x + 25) - 20(x + 25) = 0

Quadratic Equations


-22-TEACHERS FORUM

⇒ (x + 25) (x - 20) = 0

⇒ x = 20, x = -25

∴ Length of piece = 20 m

and original cost per meter = ` 20020

= ` 10

60. Some students planned a picnic. The budget for food was ` 480. But 8 of these failed to go and thus cost of food for each member increased by ` 10. How many students attended the picnic ? (2013, 2014)

Ans. Let the number of students be x

Cost for each student = 480

New Cost = 480x - 8


480x - 8

- 480x

= 10

480 ( 1x - 8

- 1x ) = 10

x - x + 8x2 - 8x

= 10480

8x2 - 8x

= 148

x2 - 8x - 384 = 0

x2 - 24x + 16x - 384 = 0

x(x - 24) + 16(x - 24) = 0

(x - 24)(x + 16) = 0

⇒x = 24, x = -16 (Rejected)

∴ No. of students = 24

61. A man drives his car on a highway where the speed limit is 60 km/hr. He has to cover a distance of 240 km at a uniform speed on this road. If he

increases his speed by 20 km/hr, he can reach his destination 1 hr earlier. What is his original speed at which he travels actually. What value do you observe through his act. (2014)

Ans. distance = 240 km

let the uniform speed = x km/h

time taken, t1 = 240

x hr

increased speed = x + 20

time taken, t2 = 240

x + 20 hr


240

x -

240x + 20

= 1

240 ( 1x

- 1

x + 20) = 1

x + 20 - xx(x + 20)

= 1

240

20

x2 + 20x =

1240

x2 + 20x - 4800 = 0 80 x -60 = - 480

x2 + 80x - 60x - 4800 = 0 80 + - 60 = 20

x(x + 80) - 60 (x + 80) = 0

(x + 80) (x - 60) = 0

x = -80 or x = 60

Since speed cannot be -ve,

his speed = 60 km/hr

His original speed is 60 km/hr, i.e., he obeys traffic rule. But if he increases his speed by 20 km/hr, he exceeds his speed limit in the highway and accident can take place.

62. A journey of 192 km from a town A to town B takes 2 hours more by an


-23-TEACHERS FORUM

ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speeds of the faster and the passenger train. (2012, 2014)

Ans. A B192 km

Let speed of passenger train be x km/h

∴speed of super fast train = (x + 16) km/h

Tpassenger = 192x

Tfast = 192x + 16


192x

- 192x + 16

= 2

192 (x + 16) - 192x = 2 (x2 + 16x)

192 (x + 16 - x) = 2(x2 + 16x)

192 x 16 = 2 (x2 + 16x)

x2 + 16x - 1536 = 0

x2 + 48x - 32x - 1536 = 0

x(x + 48) - 32(x + 48) = 0

(x - 32) (x + 48) = 0

x = 32 or - 48

Since speed can’t be negative, speed of passenger train = 32 km/h

and speed of fast train = 48 km/h.

63. The difference of square of two numbers is 180. The square of the smaller number is 8 times the large number. Find the two numbers. (2012, 2014)

Ans. Let x > y ⇒ x2 - y2 = 180 →(1)

Also given, y2 = 8x →(2)

(1) ⇒x2 - 8x - 180 = 0

x2 - 18x + 10x - 180 = 0

x(x - 18) + 10 (x - 18) = 0

(x - 18) (x + 10) = 0 ⇒x = 18

(2) ⇒ y2 = 8 x 18 = 144

⇒ y = 12

64. The time taken by Ram to cover 150km in one direction was 150 minutes more than the time in the return journey. If he returned at a speed of 10km/hr more than the speed of going. What was the speed per hour in each direction?(2014)

Ans. Let speed be x km/hr


150

x - 150

x + 10 = 150 min.

150 [ 1x -

1x + 10 ] =

15060 =

52 hr

x + 10 - xx2 + 10x =

160

x2 + 10x - 600 = 0

x2 + 30x - 20x - 600 = 0

x(x + 30) - 20 (x + 30) = 0

(x + 30) (x - 20) = 0

x = -30 or x = 20

∴ speed of going = 20km/h

and speed of return = 30km/h

65. An aeroplane left 40 minutes late due to heavy rains and in order to reach its destination, 1600km away in time, it had to increase its speed by 400 km/hr from its original speed. Find the original speed of the plane. (2012, 2014)

Quadratic Equations


-24-TEACHERS FORUM

Ans. distance = 1600 km

usual speed = x km/h

∴ normal time, t1 = 1600

x hr

increased speed = x + 400

∴ time, t2 = 1600

x + 400 According to the equation,

1600

x - 1600

x + 400 = 40 min.

1600 [ 1x -

1x + 400 ] =

23 hr

x + 400 - xx2 + 400x =

12400

x2 + 400x - 960000 = 0

x2 + 1200 x - 800x - 960000 = 0

x(x + 1200) - 800 (x + 1200) = 0

(x + 1200) (x - 800) = 0

⇒ x = 800 km/hr

[ x = -1200, rejected]

i.e., original speed = 800 km/h

66. A motor boat, whose speed is 15km/h in still water, goes 30 km down stream and comes back in a total of 4 hours and 30 minutes. Determine the speed of stream (2012)

Ans. Let speed of stream is xkm/h.

Speed in upstream = 15 - x km/h

speed in down stream = 15 + x km/h


30 15 + x

+ 30 15 - x

= 4 1 2

30 (15 - x) + 30(15 + x) (15 + x) (15 - x)

= 92

450 - 30x + 450 + 30x 225 - x2

= 92

900 225 - x2

= 92

-9x2 + 2025 - 1800 = 0

- x2 + 225 - 200 = 0

x2 - 225 + 200 = 0

x2 = 25

⇒ x = ± 5, since speed cannot be -ve

speed of stream = 5 km/hr

67. The denominator of a fraction is two more than its numerator. If the sum of the fraction and its reciprocal is 2

415 ,

find the fraction. (2012)

Ans. Let the numerator be x. Then the denominator is x + 2

Then fraction is xx + 2


x

x + 2 + x + 2

x = 2 4

15

x2 + (x + 2)2

x (x + 2) =

3415

x2 + x2 + 4x + 4x2 + 2 x

= 3415

2x2 + 4x + 4x2 + 2 x

= 3415

30x2 + 60x + 60 = 34x2 + 68x

4x2 + 8x - 60 = 0

x2 + 2x - 15 = 0

x2 + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x + 5) (x - 3) = 0

x = 5 or x = 3

x = -5 is rejected


-25-TEACHERS FORUM

∴ x = 3

Then the required fraction is xx + 2

= 35

68. Two taps together can fill a tank in 9 38

hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. (2020 - B, 2013, 2012)

Ans. Let time taken by tap of larger diameter = x hrs

∴time taken by tap of smaller diame-ter = (x + 10) hrs

∴ 1x + 10

+ 1x = 8

75 ⇒ 75 (2x + 10) = 8x (x + 10)

⇒ 8 x2 - 70x - 750 = 0

⇒ 4 x2 - 35x - 375 = 0

⇒4x2 - 60x + 25x - 375 = 0

⇒4x(x -15) + 25 (x -15) = 0

⇒ (4x + 25) (x - 15) = 0

x = 15 or x = -254

, not possible

∴time taken by tap of larger diameter

= 15 hrs

time taken by tap of smaller diameter

= 25 hrs

69. Three consecutive positive integers are taken such that the sum of the squares of the first and the product of the other two is 154. Find the integers. (2012)

Ans. Let the three consecutive integers be x, x + 1 and x + 2.


x2 + (x + 1)(x + 2) = 154

x2 + x2 + 3x + 2 - 154 = 0

2x2 + 3x - 152 = 0 19 x -16 = -304

2x2 + 19x - 16x - 152 = 0 19 + -16 = 3

x(2x + 19) - 8(2x + 19) = 0

(x - 8)(2x + 19) = 0

i.e., x = 8 or x = -192

Since x cannot be -ve integer, x = 8 and the three numbers are 8, 9 and 10.

70. By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450 km. Find the original speed of the bus. (2012)

Ans. Let the original speed = x km/hr

and distance = 450 km

∴time, t1 = 450 x

increased speed = x + 10 km/hr

∴time, t2 = 450

x + 10


450 x -

450 x + 10 = 1 1

2 hr

450(x + 10) - 450x x (x + 10) = 3

2

450x + 4500 - 450x x2 + 10x = 3

2

4500 x 2 = 3(x2 + 10x)

9000 = 3x2 + 30x

i.e., 3x2 + 30x - 9000 = 0

x2 + 10x - 3000 = 0 (splitting the middle term)

Quadratic Equations


-26-TEACHERS FORUM

x2 + 60x - 50x - 3000 = 0

x (x + 60) - 50(x + 60) = 0

(x- 50) (x + 60) = 0

x = 50 or x = -60

Since speed cannot be negative, the original speed of the bus = 50 km/hr

71. A shopkeeper buys a number of books for ` 1200. If he had bought 10 more books for the same amount, each book would have cost him ` 20 less. How many books did he buy ? (2011)

Ans. Let the no. of books purchased be ‘n’

amount paid = ` 1200

∴ cost for each book, p = 1200n

If 10 more books would have been purchased,

cost for each book, pI = 1200n + 10


p - pI = 20

i.e., 1200n - 1200

n + 10 = 20

1200 [ 1n - 1

n + 10] = 20

⇒ n +10 - nn2 + 10n = 20

1200

10n2 + 10n = 1

60 n2 + 10n = 600 n2 + 10n - 600 = 0 30 x -20 = -600

n2 + 30n - 20n - 600 = 0 30 + -20 = 10

n(n + 30) - 20(n + 30) = 0

(n + 30)(n - 20) = 0

⇒ n = -30 or n = 20

Since n cannot be negative, n = 20.

72. A person on tour has ` 360 for his expenses. If he extends his tour for four days, he has to cut down his daily expenses by ` 3. Find the original duration of the tour. (2011, 2012)

Ans. Let the original duration of the tour be x days.

Total expenditure on tour = ` 360

⇒ Expenditure per day = ` 360x

Duration of extended tour = (x + 4) days

⇒ Expenditure per day accoding to

new schedule = ` 360x + 4

Given that daily expenses are cut down by ` 3.

∴ 360x - 360

x + 4 = 3

360(x+ 4) - 360xx (x + 4)

= 3

360x + 1440 - 360xx2 + 4x = 3

1440 = 3(x2 + 4x)

3x2 + 12x - 1440 = 0

x2 + 4x - 480 = 0

x2 + 24x - 20x - 480 = 0 24 x -20 = - 480

x(x + 24) -20(x + 24) = 0 24 + -20 = 4

(x + 24) (x - 20) = 0

⇒ x - 20 = 0 or x + 24 = 0

i.e., x = 20 or x = -24

Since the number of days cannot be negative, x = 20.


-27-TEACHERS FORUM

" PRACTICE CORNER

1. Solve the following quadratic equation for x : 4x2 + 4bx - (a2 - b2) = 0 (2015)

2. Find the roots of the quadratic equation : √7 y2 - 6y - 13 √7 = 0. (2012, 2014)

3. Find the roots of the quadratic equation : x2 - x5

+ 1100

= 0 (2012, 2014)

4. Find the roots of the quadratic equation : 25 x2 - x -

35 = 0 (2012, 2014)

5. Solve for x : x2 - (√3 + 1) x + √3 = 0 (2015)

6. Solve the quadratic equation for x : x2 - 2ax - (4b2 - a2) = 0. (2015)

7. Solve the quadratic equation for x : 9x2 - 6b2x - (a4 - b4) = 0 (2015)

8. Solve for x : abx2 + (b2 - ac)x - bc = 0. (2012, 2013)

9. Solve for x : x - 1x - 2 + x - 3

x - 4 = 3 13 (x ≠ 2, 4) (2011, 2014)

10. Solve for x : (x + 47 ) + ( 7

x+4 ) = 2. (2012)

11. Solve for x: 1x + 4 - 1

x - 7 = 1130 , x ≠ -4, 7. (2012, 2013)

12. A two digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number. (2011,2012)

13. The sum of the squares of two consecutive odd natural numbers is 130. Find the numbers. (2012)

14. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is 29

20. Find the original fraction. (2015)

Ans. 1. x = a + b2

, a - b2

2. y = 13√7

, - √7 3. x = 110

, 110

4. x = -12

, x = 3 5. x = √3 , 1 6. x = a - 2b, x = a + 2b

7. x = b2 - a2

3, b2 + a2

3 8. x = -b

a or x = cb 9. x = 5 or x = 5

2

10. x = 3 11. x = 2, 1 12. 26 13. 7 and 9 14. 710

Quadratic Equations


-28-TEACHERS FORUM

EXERCISE 4.3 Topics: Solution of a quadratic

equation by quadratic formula.

2 MARKS1. If one root of the quadratic equation 3x2

– 5x + 5k = 0 is 5, find the value of k.

Ans. 3 × 52 – 5 × 5 + 5 k = 0 (2019)

⇒ 50 = -5k

⇒ k = –10

2. Solve the following equation by using the quadratic formula : (2012)

x - 1x = 3; x ≠ 0

Ans. x - 1x = 3

i.e., x2 - 1

x = 3

x2 - 3x - 1 = 0

D = b2 - 4ac

= (-3)2 - 4 x 1 x (-1) = 9 + 4 = 13

x = 3 + √13 2 or 3 - √13

2

3. Find the roots of the quadratic equation, √2 x2 + 7x + 5√2 = 0. (2011)

Ans. Here a = √2 , b = 7, c = 5√2

x = -b ± √b2 - 4ac2a

= -7 ± √49 - 402√2

= -7 ± 32√2

i.e., x = -7 + 32√2

or x = -7 - 32√2

i.e., x = -42√2

or x = -102√2

i.e., x = -2√2

or x = -5√2

4. Find the roots of the equation

1x - 1(x - 2) = 3; x ≠ 0, 2. (2011, 2014)

Ans. Given, 1x - 1

x - 2 = 3

⇒ x - 2 - xx (x - 2) = 3

⇒ - 2x2 - 2 x = 3

-2 = 3x2 - 6x

3x2 - 6x + 2 = 0

Here a = 3, b = -6, c = 2

x = -b ± √b2 - 4ac2a

= 6 ± √36 - 246 = 6 ± √12

6

= 6 ± 2√36 = 2(3 ± √3 )

6

i.e., x = 3 + √33

or x = 3 - √33

3 MARKS

5. Solve for x :

x2 - (2b - 1)x + (b2 - b - 20) = 0 (2015)

Ans. x = (2b - 1) ±√(2b - 1)2 - 4(b2 - b - 20)2

= 2b - 1 ±√4b2 - 4b + 1 - 4b2 + 4b + 802

= 2b - 1 ±√812

= 2b - 1 ±92


-29-TEACHERS FORUM

= 2b + 82

, 2b - 102

⇒ x = b + 4, b - 5

6. Solve for x : x2 + 6x - (a2 + 2a - 8) = 0.

Ans. x = -b ±√b2 - 4ac2a

(2015)

= -6 ±√36 + 4(a2 + 2a - 8)2

= -6 ±√36 + 4a2 + 8a - 322

= -6 ±√4a2 + 8a + 42

= -6 ±√(2a + 2)2

2 = -6 ±(2a + 2)

2 = 2a - 4

2 or -2a - 8

2 ∴ x = a - 2, -a - 4

7. Find the roots of the equation:

x2 - 2(a2 + b2)x + (a2 - b2)2 = 0 (2012)

Ans. Given, x2 - 2(a2 + b2)xx + (a2 - b2)2 = 0

D = b2 - 4 ac

= [-2(a2 + b2) ]2 - 4 x 1 x (a2 - b2)2

= 4(a2 + b2)2 - 4 (a2 - b2)2

= 4[ a4 + b4 + 2a2b2 - a4 - b4 + 2a2b2]

= 4[ 4 a2b2] = 16 a2b2

∴ x = -b ± √b2 - 4ac 2a

= 2(a2 + b2) ± √16a2b2 2

= 2(a2 + b2) ± 4ab2

= a2 + b2 + 2ab or a2 + b2 - 2ab

= (a + b)2 or (a - b)2

4 MARKS

8. If x = -12 is a root of the quadratic

equation 3x2 + 2kx + 3 = 0, find the value of k. (2019)

Ans. 3 ( -12 )2 + 2k (-12 ) + 3 = 0

34 - k + 3 = 0

⇒ k = 154

9. If the roots of the quadratic equation in x : (a2 + b2) x2 – 2 (ac + bd) x + (c2 + d2) = 0 are equal, prove that ad = bc.

(2019, 2017)

Ans. Roots of the given equation are equal.

∴ D = 4(ac + bd)2 – 4(a2 + b2) (c2 + d2) = 0

⇒ 4[a2c2 + b2d2 + 2acbd – a2c2 – a2d2 – b2c2 – b2d2] = 0

⇒ 4[ 2acbd – a2d2 – b2c2 ] = 0

⇒ 4(ad – bc)2 = 0

⇒ (ad – bc)2 = 0

⇒ (ad – bc) = 0

⇒ ad = bc

10. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. (2019)

Ans. x2 + px + 16 = 0 have equal roots

if D = 0

b2 - 4ac = p2 – 4 (1) (16) = 0

⇒ p2 = 64 ⇒ p = ± 8

Quadratic Equations


-30-TEACHERS FORUM

∴ x2 ± 8x + 16 = 0

(x ± 4)2 = 0

⇒ x ± 4 = 0

∴ Roots are x = –4 and x = 4

11. If one root of the quadratic equation 2x2

+ 2x + k = 0 is – 13

, then find the value of k. (2019)

Ans. Since - 13

is a root,

2(- 13)2 + 2(- 1

3) + k = 0

⇒ 29

- 23

= -k

⇒ 2 - 6

9 = -k

⇒ k = 49

12. Solve for : 1x + 1

+ 2x + 2

= 4x + 4

,

x ≠ -1, -2, -4. (2018, 2016)

Ans. x + 2 + 2(x + 1)(x + 1)(x + 2)

= 4 x + 4

3x + 4x2 + 3x + 2

= 4 x + 4

3x2 + 12x + 4x + 16 = 4x2 + 12x + 8

x2 - 4x - 8 = 0

x = 4 ± √16 + 32 2

= 4 ±4 √3 2

= 2 ± 2√3

13. A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32km upstream than to return downstream to the same spot. Find the speed of the stream. (2016)

Ans. Let x km/h be the speed of the stream

Upstream :

distance = 32 km and speed = 24 - x

∴time, t1 = 3224 - x

Downstream :

distance = 32 km and speed = 24 + x

∴time, t2 = 3224 + x

According to the problem, t1 - t2 = 1

i.e., 3224 - x

- 3224 + x

= 1

32 ( 124 - x

- 124 + x) = 1

24 + x - 24 + x576 - x2 = 1

32

64x = 576 - x2

x2 + 64x - 576 = 0

x = -64 ± √4096 + 23042

= -64 ± 802

= 8

i.e., speed of stream = 8 km/h

14. Solve for x : (2012)

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

Ans. Here a = 9, b = -9(a + b),

c = 2a2 + 5ab + 2b2

⇒ b2 - 4ac

= (-9(a+b))2 - 4 x 9 x (2a2 + 5ab + 2b2)

= 81(a + b)2 - 36(2a2 + 5ab + 2b2)

= 81(a2+ 2ab + b2) -72a2 -180ab - 72b2


-31-TEACHERS FORUM

= 81a2 + 162ab + 81b2 - 72a2 - 180ab - 72b2

= 9a2 - 18ab + 9b2

= 9(a2 - 2ab + b2)

= 9(a - b)2

⇒ √b2 - 4ac = 3(a - b)

Now, x = -b ± √b2 - 4ac2a

= 9(a + b) ± 3(a - b)

2 x 9

= 9a + 9b + 3a - 3b

18 or

9a + 9b - 3a + 3b

18

= 12a + 6b

18 or 6a + 12b

18

= 2a + b

3 or a + 2b

3

15. A motorboat whose speed is 18 km/hr in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. (2020 -S, 2018, 2014)

Ans. Let the speed of the stream be x km/hr

∴speed of the boat upstream

= (18 - x) km/hr

Speed of the boat downstream

= (18 + x) km/hr

Time taken to go upstream

= distancespeed

= 2418 - x

Time taken to go downstream = 2418 + x


2418 - x

- 2418 + x

= 1

24 [ 118 - x

- 118 + x] = 1

18 + x - 18 + x324 - x2

= 124

2x324 - x2

= 124

x2 + 48x - 324 = 0

b2 - 4ac = 482 - 4 x 1 x -324 = 3600

Now, x = -48 ±√36002

= -48 ±602

= 6 or -54

∴ Speed of the stream = 6 km/hr

16. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. (2011, 2012)

Ans. Square 1

Let side = x

Area = x2

Perimeter = 4x

Square 2

Let side = y

Area = y2

Perimeter = 4y

According to the problem, x2 + y2 = 468 →(1)

and 4x - 4y = 24

4 (x - y) = 24

x - y = 6

x = y + 6 →(2)

Substitute (2) in (1) we get,

(y + 6)2 + y2 = 468

y2 + 12y + 36 + y2 = 468

2y2 + 12y - 432 = 0

Quadratic Equations


-32-TEACHERS FORUM

y2 + 6y - 216 = 0

Using quadratic formula,

y = -6 ± √36 - 4 x 1 x -216

2 x 1

= -6 ± √9002

= -6 ± 302

= - 362

or 242

= -18 or 12

As length cannot be negative, y = 12

(2) ⇒x = 12 + 6 = 18 [From (2)]

So sides are 18 m and 12 m.

17. A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train. (2011)

Ans. Let the uniform speed be x km/hr

distance travelled = 180 km

time = 180x

If the speed is 9 km/hr more,

time taken = 180x + 9


180x - 180

x + 9 = 1

180 [ 1x - 1

x + 9] = 1

x + 9 - xx (x + 9) = 1

180

9

x2 + 9x = 1180

x2 + 9x = 1620

x2 + 9x - 1620 = 0

x = -9 ± √81 + 6480

2

= -9 ± √6561

2

= -9 ± 81

2 x =

-9 + 812

or x = -9 - 81

2 x = 36 or x = -45

Since speed cannot be negative, speed = 36 km/hr

18. The product of Tanay’s age (in years) five years ago and his age ten year later is 16. Determine Tanay’s present age. (2011, 2012)

Ans. Let the present age of Tanay be x years.

His age before five years = x - 5

His age after 10 years = x + 10


(x - 5)(x + 10) = 16

x2 + 5x - 50 = 16

x2 + 5x - 66 = 0

Here a = 1, b = 5, c = -66

x = -b ± √b2 - 4ac2a

= -5 ± √25 + 2642 =

-5 ± 172

⇒ x = -11 or 6

Since age cannot be negative, Tanay’s present age is 6 years.


-33-TEACHERS FORUM

" PRACTICE CORNER

1. Solve for x: x - 3x - 4

+ x - 5x - 6

= 103

; x ≠ 4, 6. (2013)

2. Find the roots of the equation : 12x - 3 + 1

x - 5 = 1, x ≠ 32 , 5. (2011, 2012)

3. Solve for x : a(a2 + b2) x2 + b2x - a = 0 (2011, 2012)

4. Solve for x : 9 x2 - 3(a + b) x + ab = 0 (2011, 2013)

5. Solve for x: 16x

- 1 = 15x + 1

; x ≠ 0, -1. (2014)

6. Solve for x: 2 ( 2x + 3 x - 3 ) - 25 ( x - 3

2x + 3 ) = 5 (2014)

7. Solve the following equation for x :

( 2x - 3x - 1 ) - 4 ( x - 1

2x - 3 ) = 3; x ≠ 1, x ≠32 (2012, 2014)

8. Solve for x : 3x + 1

+ 4x - 1

= 294x - 1

; x ≠ 1, -1, 14

(2015)

9. A bus travels at a certain average speed for a distance of 75 km and then travels a distance of 90 km at an average speed of 10 km/h more than the first speed. If it takes 3 hours to complete the total journey, find its first speed. (2015)

10. A two – digit number is such that the product of the digits is 35. When 18 is added to this number, the digits interchange their places. Determine the number. (2014)

Ans. 1. x = 7, 92

2. x = 8 + 3√2

2 or x =

8 - 3√22

3. x = aa2 + b2 or - 1

a

4. x = b3 , a

3

5. x = ± 4

6. x = 1, 6

7. x = 43

8. -7, 4 9. 50 km/h 10. 57

Quadratic Equations


-34-TEACHERS FORUM

EXERCISE 4.4 Topics : Nature of roots

(i) Two distinct real roots, if b2 - 4ac > 0

(ii) Two equal real roots, if b2 - 4ac = 0

(iii) No real roots, if b2 - 4ac < 0

2 MARKS

1. Find the values of p for which the quadratic equation x2 - 2px + 1 = 0 has

no real roots. (2020 - B)

Ans. b2 - 4ac < 0

(-2P)2 - 4 x 1 < 0

4p2 - 4 < 0

4 (p2 - 1) < 0

p2 - 1 < 0

⇒-1 < p < 1

2. For what values of k, the roots of the equation x2 + 4x + k = 0 are real? (2019)

Ans. Since roots of the equation x2 + 4x + k = 0 are real, b2 - 4ac ≥ 0

⇒ 16 – 4k ≥ 0

⇒ k ≤ 4

3. Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0. (2019)

Ans. 2x2 – 4x + 3 = 0 D = b2 - 4ac = 16 – 24 = –8

∴ Equation has NO real roots

4. Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3). (2019)

Ans. (x + 5)2 = 2(5x – 3)

x2 + 10x + 25 = 10x - 6

⇒ x2 + 0 + 31 = 0 D = b2 - 4ac = 0 - 4 x 1 x 31 = –124

5. For what values of k does the quadratic equation 4x2 – 12x – k = 0 have no real roots?

Ans. For no real roots, b2 - 4ac < 0 (2019) ie. 144 – 4 × 4 × (–k) < 0

⇒ 16 k < –144

k < –9

6. Find the value/s of k for which the quadratic equation 3x2 + kx + 3 = 0 has real and equal roots. (2019)

Ans. For real and equal roots, b2 - 4ac = 0 i.e., k2 – 4 × 3 × 3 = 0 ⇒ k2 = 36

⇒ k = ± 6

7. Find the value of k for which the roots of the quadratic equation 2x2 + k x + 8 = 0 will have equal value. (2017)

Ans. For equal roots, b2 - 4ac = 0

k2 – 4(2) (8) = 0

⇒ k2 = 64 ⇒ k = ± 8

8. If - 5 is a root of the quadratic equation 2x2 + px - 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k. (2016)

Ans. p(-5) = 0

i.e., 2(-5)2 + p(-5) -15 = 0

50 - 5p - 15 = 0


-35-TEACHERS FORUM

-5p = -35

⇒ p = 7

Now the given quadratic equation becomes, 7(x2 + x) + k = 0

7x2 + 7x + k = 0

b2 - 4ac = 0

49 - 4 x 7 x k = 0

28k = 49

⇒ k = 4928

= 74

9. For what value of k the equation 4x2 - 2(k + 1)x + (k + 1) = 0 has real and equal roots ? (2011, 2012)

Ans. For an equation to have real and equal roots, b2 - 4ac = 0

[-2(k + 1)]2 - 4 x 4 x (k + 1) = 0

4(k + 1)2 - 16(k + 1) = 0

4(k + 1)(k + 1 - 4) = 0

4(k + 1) (k - 3) = 0

k + 1 or k - 3 = 0

k = - 1 or +3

3 MARKS

10. Find that non-zero value of k, for which the quadratic equation kx2 + 1 - 2(k - 1)x + x2 = 0 has equal roots. Hence find the roots of the equation. (2020 - S, 2015)

Ans. The given quadratic equation can be written as

(k + 1)x2 - 2(k - 1)x + 1 = 0

For equal roots b2 - 4ac = 0

i.e., 4(k - 1)2 - 4(k + 1) = 0

4(k2 - 2k + 1) - 4k - 4 = 0

i.e.,4k2 - 12k = 0

4k(k - 3) = 0

⇒ k = 0, 3

For k = 3, the equation becomes

4x2 - 4x + 1 = 0

4x2 - 2x - 2x + 1 = 0

2x(2x - 1) - 1(2x - 1) = 0

(2x - 1) (2x - 1) = 0

i.e., x = 12

, 12

11. Find the values of k, for which the quadratic equation (k + 4) x2 + (k + 1) x + 1 = 0 has equal roots. (2020 - S)


(K + 1)2 - 4 (K + 4) x 1 = 0

K2 + 2K + 1 - 4K - 16 = 0

K2 - 2K - 15 = 0

K = 2 ± √4 + 602

= 2 ± 82

∴ K = 5, -3

12. If -4 is a root of the quadratic equation x2 + px - 4 = 0 and the quadratic equation x2 + px + k = 0 has equal roots, find the value of k. (2012)

Ans. p(x) = x2 + px - 4

since -4 is a root, p(-4) = 0

i.e., (-4)2 + p(-4) - 4 = 0

16- 4p - 4 = 0

-4p = -12

Quadratic Equations


-36-TEACHERS FORUM

p = 3

Now x2 + px + k = 0

becomes x2 + 3x + k = 0

Since it has equal root,

b2 - 4ac = 0

i.e., 32 - 4 x 1 x k = 0

9 - 4k = 0

k = 94

13. For what value(s) of k will the quadratic equation (2k + 1) x2 + 2(k + 3) x + (k + 5) = 0 have real and equal roots ? (2014)

Ans. For real and equal roots,

D = b2 - 4ac = 0

[2(k + 3)]2 - 4.(2k + 1)(k + 5) = 0

4(k2 + 6k + 9) - 4 (2k2 + 10k + k + 5) = 0

4(k2 + 6k + 9) - 8k2 - 44k - 20 = 0

4k2 + 24k + 36 - 8k2 - 44k - 20 = 0

-4k2 - 20k + 16 = 0

k2 + 5k - 4 = 0

k = -b ± √b2 - 4ac2a

= -5 ± √25 + 16 2

= -5 ± √41 2

14. If the equation (1 + m2) n2x2 + 2 mncx + (c2 - a2) = 0 has equal roots of x, prove that c2 = a2 (1 + m2). (2012, 2014)

Ans. (1 + m2) n2.x2 + 2mnc.x + (c2 - a2) = 0 has equal roots

∴b2 - 4ac = 0

{2mnc}2 - 4 {(1 + m2) n2} {c2 - a2} = 0

4m2 n2 c2 - 4 n2(1 + m2)(c2 - a2) = 0

∴ 4m2n2c2 - 4n2 (1 + m2)(c2 - a2) = 0

4n2 [m2c2 - c2 + a2 - m2c2 + m2a2] = 0

a2 + m2a2 = c2

a2(1 + m2) = c2

4 MARKS

15. Find the positive value(s) of k for which quadratic equations x2 + kx + 64 = 0 and x2 – 8x + k = 0 both will have real roots. (2016)

Ans. (i) For

x2 + kx + 64 = 0 to have real roots

b2 - 4ac ≥ 0

k2 – 256 ≥ 0 →(i)

(ii) For x2 – 8x + k = 0 to have real roots

b2 - 4ac ≥ 0

64 – 4k ≥ 0 →(ii)

For (i) and (ii) to hold simultaneously k = 16

16. If x = -2 is a root of the equation 3x2 + 7x + p = 0, find the values of k so that the roots of the equation x2 + k(4x + k - 1) + p = 0 are equal. (2015)

Ans. x = -2 is root of 3x2 - 7x + p = 0

⇒ 3(-2)2 + 7(-2) + p = 0

⇒ 12 - 14 + p = 0

⇒ p = 2

Now x2 + k(4x + k - 1) + p = 0


-37-TEACHERS FORUM

becomes x2 + k (4x + k - 1) + 2 = 0

x2 + 4kx + (k2 - k + 2) = 0

Since roots are equal b2 - 4ac = 0

i.e., 16k2 - 4(k2 - k + 2) = 0

16k2 - 4k2 + 4k - 8 = 0

12k2 + 4k - 8 = 0

⇒ 3k2 + k - 2 = 0

3k2 + 3k - 2k - 2 = 0

i.e., 3k(k + 1) - 2(k + 1) = 0

⇒ (3k - 2)(k + 1) = 0

⇒ k = 23

, -1

17. If x = -4 is a root of the equation x2 + 2x + 4p = 0, find the values of k for which the equation x2 + px (1 + 3k) + 7 (3 + 2k) = 0 has equal roots. (2015)

Ans. x = -4 is root of the equation

x2 + 2x + 4p = 0.

∴ (-4)2 + 2(-4) + 4p = 0

⇒ p = -2

Now x2 - 2(1 + 3k) x + 7(3 + 2k) = 0 has equal roots.

∴ b2 - 4ac = 0

i.e., 4(1 + 3k)2 - 28(3 + 2k) = 0

4(1 + 6k + 9k2) - 84 - 56k = 0

4 + 24k + 36k2 - 84 - 56k = 0

36k2 - 32k - 80 = 0

⇒ 9k2 - 8k - 20 = 0

⇒ (9k + 10)(k - 2) = 0

⇒ k = -109

, 2

18. If the roots of the equation (a - b)x2 + (b - c)x + (c - a) = 0 are equal,

prove that 2a = b + c. (2014)

Ans. Since roots are equal, b2 - 4ac = 0

i.e., (b - c)2 - 4 (a - b) (c - a) = 0

b2 + c2 - 2bc - 4 (ac - a2 - bc + ab) = 0

b2 + c2 - 2bc - 4ac + 4a2 + 4bc - 4ab = 0

b2 + c2 + 4a2 + 2bc - 4ab - 4 ac = 0

(2a - b - c)2 = 0

2a - b - c = 0

2a = b + c Hence proved.

19. If the roots of the equation (c2 - ab) x2 - 2(a2 - bc)x + (b2 - ac) = 0 are real and equal, then show that either a = 0 or a3 + b3 + c3 = 3abc. (2012, 2014)

Ans. Since the roots are real and equal, b2 - 4ac = 0

[-2 (a2 - bc)]2 - 4 x (c2 - ab)(b2 - ac) = 0

4(a2 - bc)2 - 4(c2b2 - ac3 - ab3 + a2bc) = 0

4[a4- 2a2bc+ b2c2 - b2c2 + ac3+ab3- a2bc] = 0

4a [a3 - 3abc + c3 + b3] = 0

⇒ 4a = 0 or a3 - 3abc + b3 + c3 = 0

⇒ a = 0 or a3 + b3 + c3 = 3abc. Hence proved.

Quadratic Equations


-38-TEACHERS FORUM

" PRACTICE CORNER

1. Find the value of k for which the following quadratic equation has two equal roots: 2 x2 + kx + 3 = 0. (2012)

2. Find the value of m so that the quadratic equation mx (x - 7) + 49 = 0 has two equal roots. (2011, 2013)

3. Find value of p such that the quadratic equation (p - 12)x2 - 2 (p - 12) x + 2 = 0 has equal roots. (2011)

4. Find the value of C for which the quadratic equation 4x2 - 2(c + 1) x + (C + 4) = 0 has equal roots. (2011, 2012)

5. If x = 3 is a root of the equation x2 - x + k = 0, find the value of p so that the roots of the equation x2 + k(2x + k + 2) + p = 0 are equal. (2015)

6. If x2 + 2kx + 4 = 0 has a root x = 2, then find the value of k.

7. Find the non-zero roots of 3x - 5x2 = 0.

8. Find the value of k for which the equation 4x2 + 4xk + 9 = 0 has equal roots.

9. If 8 is a root of the equation x2 - 10x + k = 0, then find the value of k.

10. If ½ is a root of the quadratic equation x2 + kx - 54

= 0, then find the value of k

Ans. 1. k = ± 2√6 2. m = 4 3. p = 14

4. C = 5, -3 5. p = 12 6. k = -2

7. x = 35 8. k = ± 3 9. k = 16

10. k = 2


-39-TEACHERS FORUM

1 MARK

1. Find the value(s) of k for which the roots of the quadratic equation 9x2 + 3kx + 4 = 0 are real and equal. (2020-S)

Ans. For roots to be real and equal,

b2 - 4ac = 0

9k2 – 144 = 0

k2 = 1449 = 16

⇒ k = ± 4

2. If -2 is a root of the quadratic equation 3x2 - 5x + k = 0, then the value of k is _______. (2020-S)

Ans. 3 x (-2)2 - 5 (-2) + k = 0

⇒ k = -22

3. The roots of the equation, x2 + bx + c = 0 are equal if _______. (2020-B)

Ans. b2 = 4c

4. Value of the roots of the quadratic equation, x2 - x - 6 = 0 are ________.

Ans. x2 - 3x + 2x - 6 (2020-B)

= x(x - 3) + 2 (x - 3)

= (x - 3) (x + 2)

Roots are 3 and -2

5. If quadratic equation 3x2 – 4x + k = 0 has equal roots, then the value of k is _______. (2020-B)

Ans. 43

6. Find the value of k for which the roots of the quadratic equation (k – 5)x2 + 2(k – 5) x + 2 = 0 are equal. (2018)

Ans. Since roots are equal, b2 - 4ac = 0

⇒ [2 (k - 5)]2 - 4 x (k - 5) x 2 = 0

⇒ 4(k - 5)2 - 8 (k - 5) = 0

⇒ 4 (k - 5) [k - 5 - 2] = 0

⇒ 4 (k - 5) (k - 7) = 0

⇒ k = 5, k = 7

But k ≠ 5 ∴ k = 7

7. If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k. (2018)

Ans. x = 3 is one root of the equation

∴9 - 6k - 6 = 0

-6k = -3 ⇒ k = 12

8. Find the value of p for which the quadratic equation px (x – 3) + 9 = 0 has equal roots. (2018)


i.e., 9p2 – 36p = 0

⇒ 9p (p - 4) = 0

⇒ p = 4

9. If 12 is a root of the equation x2 + kx –

54

= 0, then the value of k is _________.

Ans. Given, (12 )2 + k (

12 ) -

54 = 0

⇒14 +

k2 -

54 = 0

⇒ 1 + 2k - 5

4 = 0

⇒ k = 2

10 Find the nature of roots of the quadratic equation 3x2 + 2x + 5 = 0. (2015)

Ans. b2 - 4ac = (2)2 - 4 x 3 x 5 = 4 - 60 = - 56

Since b2 - 4ac < 0, the quadratic equation has no real roots.

Quadratic Equations


-40-TEACHERS FORUM

Multiple Choice Questions :

1. The value (s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is (2020-S)

(a) 4 (b) ±4 (c) -4 (d) 0

2. The roots of the quadratic equation x2 - 0.04 = 0 are (2020 - S)

(a) ± 0.2 (b) ± 0.02

(c) 0.4 (d) 2

3. The quadratic equation x2 - 4x + k = 0 has distinct real roots if (2020-S)

(a) k = 4 (b) k > 4

(c) k = 16 (d) k < 4

4. The discriminant of the quadratic equation 2x2 - 4x + 3 = 0 is (2020-B)

(a) -8 (b) 10 (c) 8 (d) 2√2

5. The value of l for which (x2 + 4x + l) is a perfect square, is (2020 - S)

(a) 16 (b) 9 (c) 1 (d) 4

6. The value of k for which roots of the quadratic equation kx2 + 2x + 3 = 0 are equal is :

(a) 13 (b) - 13 (c) 3 (d) -3

7. ax2 + bx + c = 0, a > 0, b = 0, c > 0 has

(a) two equal real roots

(b) one real root

(c) two distinct real roots

(d) no real roots

8. If no roots of the equation x2 - px + 1 = 0 is real, then

(a) p > 2 (b) p < -2

(c) p = 2 (d) -2 < p < 2

9. If the roots of the equation 12x2 + mx + 5 = 0 are real and different then m is equal to :

(a) 8√15 (b) 2√15

(c) 4√15 (d) 10√5

10. If the discriminant of the equation 6x2 - bx + 2 = 0 is 1, then the value of ‘b’ is :

(a) 7 (b) -7 (c) ± 7 (d) ±√7

11. The roots of x2 - 2x - (r2 - 1) = 0 are :

(a) 1 - r, -r - 1 (b) 1 - r, r + 1

(c) 1, r (d) 1 - r, r

Ans. 1. (b) 2. (a) 3. (d)

4. (a) 5. (d) 6. (a)

7. (d) 8. (a) 9. (c)

10. (c) 11. (b)

Hint :

1. b2 - 4ac = 0

k2 - 4 x 2 x 2 = 0

k2 - 16 = 0

k2 = 16

k = ± 4

3. b2 - 4ac > 0

16 - 4k > 0

- 4k < -16

k < 4

4. b2 - 4ac = 16 - 4 x 2 x 3 = -8


-41-TEACHERS FORUM

CASE STUDY

1. Quadratic equations arise in several situations in the world around us and in different fields of mathematics.

(a) The standard form of a quadratic equation is:

(i) ax2 + bx + c = 0, a ≠ 0 (ii) ax2 + bx + c = 0, b ≠ 0

(iii) ax2 - bx + c = 0, a ≠ 0 (iv) ax2 - bx + c = 0, b ≠ 0

(b) Which of the following is a quadratic equations:

(i) (x – 2)2 + 1 = x2 – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)

(iii) x (2x + 3) = x2 + 1 (iv) (x – 2)(x + 1) = (x – 1)(x + 3)

(c) A quadratic equation has no real roots, if :

(i) b2 - 4ac < 0 (ii) b2 - 4ac > 0 (iii) b2 - 4ac = 0 (iv) b2 = 4ac

(d) If 8 is a root of the equation x2 - 10x + k = 0, then find the value of k.

(i) -4 (ii) 0 (iii) 16 (iv) 4

(e) Choose the correct representation of the following situations in the form of quadratic equations :

‘The product of two consecutive positive integers is 306’.

(i) x2 + (x + 1)2 = 306 (ii) x x (x + 1) = 306

(iii) x2 x (x + 2)2 = 306 (iv) x x (x + 2) = 306

Ans. (a) (i) ax2 + bx + c = 0, a ≠ 0

(b) (iii) x (2x + 3) = x2 + 1

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

(d) (iii) 16

(e) (ii) x x (x + 1) = 306

Quadratic Equations


-42-TEACHERS FORUM

" SELF ASSESSMENT TEST : 1

1. Value(s) of k for which the quadratic equation 2x2 - kx + k = 0 has equal roots is/are

(a) 0 (b) 4 (c) 8 (d) 0, 8

2. The linear factors of the quadratic equation x2 + kx + 1 = 0 are

(a) k ≥ 2 (b) k ≤ 2 (c) k ≥ - 2 (d) 2 ≤ k ≤ - 2

3. If one root of the quadratic equation ax2 + bx + c = 0 is the reciprocal of the other, then

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

4. (x2 + 1)2 - x2 = 0 has

(a) four real roots (b) two real roots (c) no real roots (d) one real root

5. If x2 + y2 = 25, xy = 12, then x is

(a) (3, 4) (b) (3,- 3) (c) (± 3, ±4) (d) (3,- 3)

6. Solve the quadratic equation for x : 4√3 x2 + 5x - 2√3 = 0 (2013, 2012)

7. Find the roots of the quadratic equation : (x + 3)(x - 1) = 3(x - 13) (2012)

8. Find the roots of the quadratic equation: 25

x2 - x - 35

= 0 (2012)

9. Solve the quadratic equation for x : x2 - 2ax - (4b2 - a2) = 0 (2015)

10. Find the nature of the roots of the quadratic equation : 13√3 x2 + 10x + √3 = 0 (2012)

11. Solve for x : 2x

x - 3 + 1

2x + 3 +

3x + 9(x - 3)(2x + 3)

= 0, x ≠ 3, -32 (2016)

12. Solve for x : x2 + ( aa + b + a + b

a ) x + 1 = 0 (2016)

13. Solve for x : ( 2xx - 5 )2

+ 5 ( 2xx - 5 ) - 24 = 0, x ≠ 5 (2016)

14. Solve for x : 9x2 - 9(a + b)x + 2a2 + 5ab + 2b2 = 0 (2016)

15. If (x2 + y2) (a2 + b2) = (ax + by)2, prove that xa = y

b (2014)


-43-TEACHERS FORUM


1. If the equation (m2 + n2) x2 − 2(mp + nq) x + p2 + q2 = 0 has equal roots, then

(a) mp = nq (b) mq = np (c) mn = pq (d) mq = √np

2. The quadratic equation 2x2 − √5 x + 1 = 0 has

(a) two distinct real roots (b) two equal real roots

(c) no real roots (d) more than 2 real roots

3. The real roots of the equation x2/3 + x1/3 - 2 = 0 are

(a) 1, 8 (b) -1, - 8 (c) -1, 8 (d) 1, - 8

4. If x2 + y2 = 25, xy = 12, then x is

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

5. The quadratic equation 2x2 - 3√2 x + 94

= 0 has



6. Find the roots of the quadratic equation √2 x2 + 7x + 5√2 = 0 (2017)7. Solve for x : √3 x2 + 10x + 7√3 = 0 (2017)

8. Solve the quadratic equation (x - 1)2 - 5(x - 1) - 6 = 0 (2015)

9. Solve for x : x2 + 6x - (a2 + 2a - 8) (2015)

10. Find the value of k for which the quadratic equation (2015) (k - 2) x2 + 2(2k - 3)x + (5k - 6) = 0) has equal roots.

11. Solve for x : 3

x + 1 + 4

x - 1 = 29

4x - 1 ; x ≠ -1, 1, 14 (2015)

12. If (1 + m2) x2 + 2mcx + (c2 - a2) = 0 has equal roots, prove that c2 = a2 (1 + m2) (2015)

13. Find the roots of the quadratic equation √3 x2 - 2x - √3 = 0 [2012, 2011]

14. Find the value of k , for which one root of the quadratic equation kx2 - 14x + 8 = 0 is six times the other. [2016]

15. If one root of the quadratic equation 6x2 - x - k = 0 is 23

, then find the value of k. [2017]

Quadratic Equations


-44-TEACHERS FORUM


1. The quadratic equation x2 + 3x + 2√2 = 0 has



2. The quadratic equation 3x2 + 4√3 x + 4 has



3. Solve for x : 2x2 + 6√3 x - 60 = 0 (2015)

4. Solve the equation using method of factorisation : 4x - 3 = 5

2x + 3 , x ≠ 0, -32 (2012)

5. Find the roots of the quadratic equation 2x2 - x - 6 = 0.

6. Find the values of k for which the quadratic equation 9x2 - 3kx + k = 0 has equal roots.

7. Find the value(s) of k, if the quadratic equation 3x2 - k √3 x + 4 = 0 has equal roots. (SQP 2018)

8. A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train. (SQP 2018)

9. Solve for x : 9x2 - 6ax + (a2 - b2) = 0 (2012)

10. Solve for x: xx + 1 + x + 1

x = 3415 (x ≠0, -1) (2011, 2014)

QUADRATIC EQUATIONS 4 - Books+App

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