CBSE 10th Maths Exam 2020: Important MCQs from Chapter 4 Quadratic Equations with Detailed Solutions

CBSE 10th Maths Exam 2020: Important MCQs from Chapter 4 Quadratic Equations with Detailed Solutions

Free PDF Download of CBSE Class 10 Maths Chapter 12 Areas Related to Circles Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Areas Related to Circles MCQs with Answers to know their preparation level.

Quadratic Equations Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
Find the roots of the equation x2 - 3x - m (m + 3) = 0, where m is a constant. (2011OD)
Solution:
x² - 3x - m(m + 3) = 0
D = b² - 4ac
D = (- 3)² - 4(1) [-m(m + 3)]
= 9 + 4m (m + 3)
= 4m² + 12m + 9 = (2m + 3)²

∴ x = m + 3 or -m

Question 2.
If 1 is a root of the equations ay² + ay + 3 = 0 and y² + y + b = 0, then find the value of ab. (2012D)
Solution:
ay² + ay + 3 = 0
a(1)² + a(1) + 3 = 0
2a = -3
a =

y² + y + b = 0
1² + 1 + b = 0
b = -2
∴ ab = = 3

Question 4.
If the quadratic equation px\frac{1}{2} - 2 px + 15 = 0 has two equal roots, then find the value of p. (2015OD)
Solution:
The given quadratic equation can be written as px\frac{1}{2} - 2 px + 15 = 0
Here a = p, b = - 2 p, c = 15
For equal roots, D = 0
D = b² - 4ac - 0 ...[∵ Equal roots
0 = (-2 p)² - 4 × p × 15
0 = 4 × 5p² - 60p
0 = 20p² - 60p => 20p² = 60p
p = = 3 ∴ p = 3

Quadratic Equations Class 10 Important Questions Short Answer-I (2 Marks)

Question 5.
Find the value of p so that the quadratic equation px(x - 3) + 9 = 0 has two equal roots. (2011D, 2014OD)
Solution:
We have, px (x - 3) + 9 = 0
px² - 3px + 9 = 0 Here a = p, b = -3p,
D = 0
b² - 4ac = 0 ⇒(-3p)² - 4(p)(9) = 0
⇒9p² - 36p = 0
⇒9p (p - 4) = 0
⇒9p = 0 or p - 4= 0
p = 0 (rejected) or p = 4
∴ p = 4 .....(∵ Coeff. of x² cannot be zero

Question 6.
Find the roots of 4x² + 3x + 5 = 0 by the method of completing the squares. (2011D)
Solution:
Here 4x² + 3x + 5 = 0


But cannot be negative for any real value of x.

Question 7.
Find the value of m so that the quadratic equation mx (x - 7) + 49 = 0 has two equal roots. (2011OD)
Solution:
We have, mx (x - 7) + 49 = 0
mx² - 7mx + 49 = 0
Here, a = m, b = - 7m, c = 49
D = b² - 4ac = 0 ...[For equal roots
⇒(-7m)² - 4(m) (49) = 0
⇒49m² - 4m (49) = 0
⇒49m (m - 4) = 0
⇒49m = 0 or m - 4 = 0
m = 0 (rejected) or m = 4
∴ m = 4

Question 8.
Solve for x:
36x² - 12ax + (a² - b²) = 0 (2011OD)
Solution:
We have, 36x² - 12ax + (a² - b²) = 0
⇒(36x² - 12ax + a²) - b² = 0
⇒[(6x)² - 2(6x)(a) + (a)²] - b² = 0
⇒(6x - a)² - (b)² = 0 ...[∵ x² - 2xy + y² = (x - y)²
⇒(6x - a + b) (6x - a - b) = 0 „[∵ x² - y² = (x + y)(x - y)
⇒6x - a + b = 0 or 6x - a - b = 0
⇒6x = a - b or 6x = a + b
⇒x = or

Question 9.
Find the value(s) of k so that the quadratic equation x² - 4kx + k = 0 has equal roots. (2012D)
Solution:
We have, x² - 4kx + k = 0
Here a = 1, b = -4k:, c = k D = 0 ...[Since, Equal roots
As b² - 4ac = 0
⇒(-4k)² - 4(1) (k) = 0
⇒16k² - 4k = 0 ⇒4k(4k - 1) = 0
⇒4k = 0 or 4k - 1 = 0
k = 0 (rejected) or 4k = 1
∴ k =

Question 10.
Find the value of k for which the equation x² + k(2x + k - 1) + 2 = 0 has real and equal roots. (2017D)
Solution:
We have, x² + k(2x + k - 1) + 2 = 0
x² + 2kx + k² - k + 2 = 0
Here a = 1, b = 2k, c = k² - k + 2
D = 0 ...[real and equal roots
∴ b² - 4ac = 0
⇒(2k)² - 4 × 1(k² - k + 2) = 0
⇒4k² - 4 (k² - k + 2) = 0
⇒4(k² - k² + k - 2) = 0 ⇒4(k - 2) = 0
⇒k - 2 = 0 ⇒k = 2

Question 11.
Find the value of p for which the roots of the equation px(x - 2) + 6 = 0, are equal. (2012OD)
Solution:
We have , px(x - 2) + 6 = 0
px² - 2px + 6 = 0, p ≠ 0
Two equal roots ...[Given
b² - 4ac = 0 ....[a = p, b = -2p, c = 6
∴ (-2p)² - 4(p)(6) = 0
4p² - 24p = 0 ⇒4p(p - 6) = 0
4p = 0 or p - 6 = 0
p = 0 (rejected) or p = 6
Since p cannot be equal to 0.
...[Standard form of a quad. eq. ax² + bx + c = 0, a ≠ 0
∴ P = 6

Question 12.
Solve the following quadratic equation for x: 4 x² + 5x - 2 = 0 (2013D)
Solution:
4 x² + 5x - 2 = 0
4 x² + 8x - 3x - 2 =0
4x( x + 2) - ( x + 2) = 0
( x + 2)(4x - ) = 0
x + 2 = 0 or 4x - = 0

Question 13.
Solve the following for x: x² + 7x + 5 = 0 (2017D )
Solution:
x² + 7x + 5 = 0
⇒ x² + 5x + 2x + 5 = 0
x( x + 5) + ( x + 5) = 0
( x + 5)(x + ) = 0

Question 14.
Solve the quadratic equation 2x² + ax - a² = 0 for x. (2014D)
Solution:
We have, 2x² + ax - a² = 0
2x² + 2ax - ax - a² = 0
2x(x + a) - a(x + a) = 0
(x + a) (2x - a) = 0
x + a = 0 or 2x - a = 0
∴ x = -a or x =
Alternatively:
First calculate D = b² - 4ac
Then apply x =
We get x = -a, x =

Question 15.
Find the values of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots. (2014OD)
Solution:
Given: 4x² + px + 3 = 0
Here a = 4, b = p. (= 3 ... [Equal roots
D = 0 (Equal roots)
As b² - 4ac = 0
∴ (p)² - 4(4)(3) = 0
= p² - 48 = 0 ⇒p² = 48
∴ p =

Question 16.
Solve the following quadratic equation for x: 4x²â€“ 4a²x + (a⁴ - b⁴) = 0. (2015D)
Solution:
The given quadratic equation can be written as,
4x² - 4a²x + (a4⁴ - b⁴) = 0
(4x² - 4a²x + a⁴) - b⁴ = 0
or (2x - a²)² - (b²)² = 0
⇒(2x - a² + b²) (2x - a² - b²) = 0
⇒(2x - a² + b²) = 0 or (2x - a² - b²) = 0
∴ x = or x =

Question 17.
Solve the following quadratic equation for x: 9x² - 6b²x - (a⁴ - b⁴) = 0 (2015D)
Solution:
The given quadratic equation can be written as
(9x² - 6b²x + b⁴) - a⁴ = 0
⇒(3x - b²)² - (a²)² = 0
⇒(3x - b² + a²) (3x - b² - a²) = 0 ...[:: x² - y² = (x + y) (x - y)
⇒3x - b² + a² = 0 or 3x - b² - a² = 0
⇒3x = b² - a² or 3x = b² + a²

Question 18.
Solve the following quadratic equation for x: 4x² + 4bx - (a² - b²) = 0 (20150D)
Solution: The given quadratic equation can be written as
4x² + 4bx + b² - a²2 = 0
⇒(2x + b)² - (a)² = 0
⇒(2x + b + a) (2x + b - a) = 0 ...[x² - y² = (x + y)(x - y)
⇒(2x + b + a) = 0 or (2x + b - a) = 0
⇒2x = -(a + b) or 2x = (a - b)

Question 19.
Solve the following quadratic equation for x: x² - 2ax - (4b² - a²) = 0) (2015OD)
Solution:
Given quadratic equation can be written as
x² - 2ax - 4b² + a² = 0.
(x² - 2ax + a²) - 4b² = 0 or (x - a)² - (2b)² = 0
As we know,
[a² - b² = (a + b)(a - b)]
∴ (x - a + 2b) (x - a - 2b) = 0
⇒x - a + 2b = 0 or x - a - 2b = 0
⇒x = a - 2b or x = a + 2b
⇒x = a - 2b and x = a + 2b

Question 20.
If x = and x = -3 are roots of the quadratic equation ax² + 7x + b = 0, find the values of a and b. (2016D)
Solution:
We have, ax² + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
Now, α = and β = -3 ... [Given

Question 21.
Find the value of p, for which one root of the quadratic equation px² - 14x + 8 = 0 is 6 times the other. (20170D)
Solution:
Given equation is px² - 14x + 8 = 0.
Here a = p b = -14 c = 8
Let roots be a and 6α.

Question 22.
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. (2016OD)
Solution:
We have, 2x² + px - 15 =0
Since (-5) is a root of the given equation
∴ 2(-5)² + p(-5) - 15 = 0
⇒2(25) - 5p - 15 = 0
⇒50 - 15 = 5p
⇒35 = 5p ⇒p = 7 ...(i)
Now, p(x² + x) + k ⇒px² + px + k = 0
7x² + 7x + k = 0 ...[From (i)
Here, a = 7, b = 7, c = k
D = 0 ...[Roots are equal
b² - 4ac = 0
⇒(7)² - 4(7)k = 0 ⇒49 - 28k = 0
⇒49 = 28k ∴ k =

Question 23.
Solve for x: + x = 13. (20160D)
Solution:
+ x = 13 ...(i)
⇒ = 13 - x
⇒( )² = (13 - x)² ...[Squaring both sides
⇒2x + 9 = 169 + x²2 - 26x
⇒0 = 169 + x²2 - 26x - 2x - 9
⇒x² - 28x + 160 = 0
⇒x² - 20x - 8x + 160 = 0
⇒x(x - 20) - 8(x - 20) = 0
⇒(x - 20) (x - 8) = 0
⇒x - 8 = 0 or x - 20 = 0
⇒x = 8 or x = 20
Checking, When x = 8 in (i)
+ 8 = 13
+ 8 = 13
5 + 8 = 13 ⇒13 = 13 ...[True
∴ x = 8 is the solution.
Checking, When x = 20 in (i),
+ 20 = 13 ... [From (1)
+ 20 = 13
7 + 20 ≠ 13 ...[False
∴ x = 20 is not a solution.
Therefore, x = 8 is the only solution.

Question 24.
Solve for x: - (2x - 7) = 0 (20160D)
Solution:
- (2x - 7) = 0...(i)
⇒ = 2x - 7
⇒( )² = (2x - 7)² ...[Squaring both sides
⇒6x + 7 = 4x² - 28x + 49
⇒0 = 4x² - 28x - 6x - 7 + 49
⇒4x² - 34x + 42 = 0
⇒2x² - 17x + 21 = 0 ...[Dividing by 2
⇒2x² - 14x - 3x + 21 = 1
⇒2x(x - 7) - 3(x - 7) = 0
⇒(x - 7) (2x - 3) = 0
⇒x - 7 = 0 or 2x - 3 = 0 =
x= 7 or x =
Checking: When x = 7 in (i),
- [2(7) - 7] = 0
- (14 - 7) = 0
7 - 7 = 0 ... [True
Checking: When x = in (i),
= 0
- (3 - 7) = 0
- (-4) = 0
4 + 4 ≠ 0
∴ x = 7 is the only solution.

Quadratic Equations Class 10 Important Questions Short Answer-II (3 Marks)

Question 25.
ind the roots of the following quadratic equation: 2 x² - 5x + = 0 (2011D)
Solution:
We have, 2 x² - 5x + = 0
Here, a = 2 , h = -5, c =
D = b² - 4ac
∴ D = (-5)² - 4 (2 )( )
= 25 - 24 = 1

Question 26.
Solve for x: 4x² - 4ax + (a² - b²) = 0 (2011OD)
Solution:
4x² - 4ax + (a² - b²) = 0
⇒[4x² - 4ax + a²] - b² = 0
⇒[(2x)² - 2(2x)(a) + (a)²] - b² = 0
⇒(2x - a)² - (b)² = 0
⇒(2x - a + b) (2x - a - b) = 0
⇒2x - a + b = 1 or 2x - a - b = 0
2x = a - b or 2x = a + b
∴ x = or x =

Question 27.
Solve for x: 3x²} - 2 x + 2 = 0 (2012D)
Solution:
3x²} - 2 x + 2 = 0
⇒3x² - x - x + 2 = 0
⇒ x ( x - - ( x - ) = 0
⇒( x - )( x - ) = 0
⇒ x - = 0 ⇒x =
∴ x = ....[

Question 28.
Find the value(s) of k so that the quadratic equation 2x² + kx + 3 = 0 has equal roots. (2012D)
Solution:
Given: 2x² + kx + 3 = 0
Here a = 2, b = k, c= 3
D = 0 ... [Since roots are equal
As b² - 4ac = 0 ∴ K² - 4(2)(3) = 0
K² - 24 = 0 or k² = 24
∴ k =

Question 29.
Find the value(s) of k so that the quadratic equation 3x² - 2kx + 12 = 0 has equal roots. (2012D)
Solution:
Given: 3x² - 2kx + 12 = 0
Here a = 3, b = -2k, c = 12
D = 0 ... [Since roots are equal As
b² - 4ac = 0
∴ (-2k)² - 4(3) (12) = 0
⇒4k² - 144 = 0 ⇒k² = = 36
∴ k =

Question 30.
Solve the following quadratic equation for x: x² - 4ax - b² + 4a² = 0 (2012OD)
Solution:
We have, x² - 4ax - b² + 4a² = 0
⇒x² - 4ax + 4a² - b\frac{144}{4} = 0
⇒[(x)\frac{144}{4} - 2(x)(2a) + (2a)\frac{144}{4}] - (b)² = 0
(x - 2a)² - (b)² = 0
(x - 2a + b) (x - 2a - b) = 0
x - 2a + b = 0 or x - 2a - b = 0
∴ x = 2a - b or x = 2a + b

Question 31.
Find the value of k for which the roots of the equation kx(3x - 4) + 4 = 0, are equal. (20120D)
Solution:
We have, kx(3x - 4) + 4 = 0
3kx² - 4kx + 4 = 0
Here a = 3k, b = -4k, c = 4
D = 0 ... [Since roots are equal
b² - 4ac = 0
∴ (-4k)² - 4(3k) (4) = 0
16k² - 48k = 0
16k (k - 3) = 0
16k = 0 or k - 3 = 0
k = 0 or k = 3
...[Rejecting k = 0, as coeff. of x² cannot be zero
∴ k = 3

Question 32.
Find the value of m for which the roots of the equation. mx (6x + 10) + 25 = 0, are equal. (2012OD)
Solution:
We have, mx(6x + 10) + 25 = 0
6mx² + 10mx + 25 = 0
Here a = 6m, b = 10m, c = 25
D = 0 ... Since roots are equal
b² - 4ac = 0
∴ (10m)² - 4(6m) (25) = 0
100m² - 600m = 0 ⇒100m (m - 6) = 0
100m = 0 or m - 6 = 0
m = 0 or m = 6
...[Rejecting m = 0, as coeff. of x² cannot be zero
∴ m = 6

Question 33.
For what value of k, the roots of the quadratic equation kx(x - 2 ) + 10 = 0, are equal? (2013D)
Solution:
We have, kx(x - 2 ) + 10 = 0
kx² - 2 kx + 10 = 0
Here a = k, b= -2 k, c= 10
D = 0 ...[∵ Roots are equal
As b² - 4ac = 0
∴ (-2 k)² - 4(k)(10) = 0
20k² - 40k = 0
⇒20k(k - 2) = 0
∴ 20k = 0 or k - 2 = 0
k = 0 (rejected) or k = 2
...[∵ Coeff. of x² cannot be zero
∴ k= 2

Question 34.
For what values of k, the roots of the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 are equal? (2013D)
Solution:
We have, (k + 4) x² + (k + 1) x + 1 = 0
Here, a = k + 4, b = k + 1, c = 1
D =0 ...[∵ Roots are equal
b² - 4ac = 0
∴ (k + 1)² - 4(k + 4)(1) = 0
k² + 2k + 1 - 4k - 16 = 0
k² - 2k - 15 = 0
k² - 5k + 3k - 15 = 0
k(k - 5) + 3(k - 5) = 0
(k - 5)(k + 3) = 0
k - 5 = 0 or k + 3= 0
k = 5 or k = -3
∴ k = 5 and -3

Question 35.
For what value of k, are the roots of the quadratic equation: (k - 12)x² + 2(k - 12)x + 2 = 0 equal? (2013OD)
Solution:
We have, (k - 12)x² + 2(k - 12)x + 2 = 0
The given quadratic equation will have equal roots if D = 0 ⇒b² - 4ac = 0
Here a = (k - 12), b = 2(k - 12), c = 2
b² - 4ac = 0
0 = 4(k - 12)² - 4 × (k - 12) × 2
0 = (k - 12)[4(k - 12) - 4 × 2]
0 = (k - 12) 4[k - 12 - 2]
0 = 4(k - 12) (k - 14)
∴ 4(k - 12)(k - 14) = 0
∴ k = 12 (rejected) or k = 14
But k cannot be equal to 12 because in that case the given equation will imply 2 = 0 which is not true.
∴ k = 14

Question 36.
For what value of k, are the roots of the quadratic equation y² + k² = 2 (k + 1)y equal? (2013OD)
Solution:
y² + k² = 2(k + 1)y
y² - 2(k + 1)y + k² = 0
Here a = 1, b = -2(k + 1), c = k²
D = 0 ... [Roots are equal
b² - 4ac = 0
∴ [-2(k + 1)]² - 4 × (1) × (k²) = 0
⇒4(k² + 2k + 1) - 4k² = 0
⇒4k² + 8k + 4 - 4k² = 0
⇒8k + 4 = 0
⇒8k = -4 ∴ k =

Question 37.
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. (2015D)
Solution:
The given quadratic equation can be written as
kx² + x² - 2(k - 1)x + 1 = 0
(k + 1) x² - 2 (k - 1) x + 1 = 0 ...(i)
Here, a = (k + 1), b = -2(k - 1), c = 1
For equal roots, D = 0
D = b² - 4ac
⇒(-2(k - 1)]² - 4 × (k + 1) × 1 = 0
⇒4(k − 1)² - 4(k + 1) = 0
⇒4k² + 4 - 8k - 4k - 4 = 0
⇒4k² - 12k = 0 ⇒4k(k - 3) = 0
k = 3 or k = 0 (rejected)
∴ k = 3
Putting k = 3 put in equation (i), we get
⇒4x² - 4x + 1 = 0
⇒4x² - 2x - 2x + 1 = 0
⇒2x(2x - 1) - 1(2x - 1) = 0
⇒(2x - 1) (2x - 1) = 0
⇒2x - 1 = 0 or 2x - 1 = 0
⇒x = or x =
Roots are

Question 38.
Find that value of p for which the quadratic equation (p + 1)x² - 6(p + 1)x + 3 (p + 9) = 0, p ≠ -1 had equal roots. (2015D(
Solution:
For the given quadratic equation to have equal roots, D = 0
Here a = (p + 1), b = -6(p + 1), c = 3(p + 9)
D = b² - 4ac
⇒[-6(p + 1)]² - 4(p + 1).3 (p + 9) = 0
⇒36(p + 1)² - 12(p + 1) (p + 9) = 0
⇒12(p + 1) (3p + 3 - p - 9) = 0
⇒12(p + 1)(2p - 6) = 0
⇒24(p + 1)(p - 3) = 0
⇒p + 1 = 0 or p - 3 = 0
⇒p = -1 (rejected) or p = 3
∴ p = 3

Question 39.
Solve for x: x² - 2 x - 2 = 0 (2015OD)
Solution:
The given quadratic equation can be written as
x² - 2 x - 2 = 0
Here, a = , b = -2 , c= -2
D = b² - 4ac

Question 40.
Solve for x: 2x² + 6 x - 60 = 0 (2015OD)
Solution:
Given equation can be written as
2(x² + 3 x - 30) = 0
x² + 3 x - 30 = 0
Here, a = 1, b = 3 , C = -30
D = b² - 4ac

Question 41.
If the roots of the quadratic equation (a - b)x² + (b - c)x + (c - a) = 0 are equal, prove that 2a = b + c. (2016OD)
Solution:
Here’a’ = a - b, ‘b’ = b - c, ‘c’ = c - a
D = 0 ....[Roots are equal
b² - 4ac = 0
⇒(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
⇒4a² + b2² + c² - 4ab + 2bc - 4ac = 0
⇒(-2a)² + (b)² + (c)2² + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0
⇒[(-2a) + (b) + (c)]² = 0
....[∵ x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²
Taking square-root on both sides
-2a + b + c = 0
⇒b + c = 2a ∴ 2a = b + c

Question 42.
Solve the equation , for x. (2014D)
Solution:

⇒5x = (2x + 3) (4 - 3x)
⇒5x = 8x - 6x² + 12 - 9x
⇒5x - 8x + 6x² - 12 + 9x = 0
⇒6x² + 6x - 12 = 0
⇒x² + x - 2 = 0 ...[Dividing by 6
⇒x² + 2x - x - 2 = 0
⇒x(x + 2) - 1(x + 2) = 0
⇒x - 1 = 0 or x + 2 = 0
∴ x = 1 or x = -2

Question 43.
Solve the equation ; x ≠ -1, x ≠ for x. (2014D)
Solution:

⇒2(2x + 2) = (5 - x)(3x - 1)
⇒4x + 4 = 15x - 5 - 3x² + x
⇒4x + 4 - 15x + 5 + 3x² - x = 0
⇒3x² - 12x + 9 = 0
⇒x² - 4x + 3 = 0 ...[Dividing by 3
⇒x² - 3x - x + 3 = 0
⇒x(x - 3) - 1(x - 3) = 0
⇒(x - 1) (x - 3) = 0
⇒x - 1 = 0 or x - 3 = 0
∴ x = 1 or x = 3

Question 44.
Solve the equation ; x ≠ -3, -1, for x. (2014D)
Solution:

⇒5(x + 3) = (11 - x) (x + 1)
⇒5x + 15 = 11x + 11 - x² - x
⇒5x + 15 - 11x - 11 + x² + x = 0
⇒x² - 5x + 4 = 0
⇒x² - 4x - x + 4 = 0
⇒x(x - 4) - 1(x - 4) = 0
⇒(x - 1) (x - 4) = 0
⇒x - 1=0 or x - 4 = 0
⇒x= 1 or x = 4

Question 45.
Solve for x: = 0, x ≠ 3,
Solution:

⇒4x² + 6x + x - 3 + 3x + 9 = 0
⇒4x² + 10x + 6 = 0
⇒2x² + 5x + 3 = 0 ...[Dividing both sides by 2
⇒2x² + 3x + 2x + 3 = 0
⇒x(2x + 3) + 1(2x + 3) = 0
⇒(2x + 3) (x + 1) = 0
⇒2x + 3 = 0 or x + 1 = 0
⇒x= or x = -1
But, x ≠ 2 ...[Given
∴ x= -1 is the only solution.

Question 46.
Solve for x: ; x ≠1, -2, 2 (2016D)
Solution:

⇒(2x² + 4)(x - 2) = (2x - 11)(x² + x - 2)
⇒2x³ - 4x² + 4x - 8 = 2x³ + 2x² - 4x - 11x² - 11x + 22
⇒2x³ - 4x² + 4x - 8 - 2x³ - 2x² + 4x + 11x² + 11x - 22 = 0
⇒5x² + 19x - 30 = 0
⇒5x² + 25x - 6x - 30 = 0
⇒5x(x + 5) - 6(x + 5) = 0
⇒(x + 5) (5x - 6) = 0
⇒x + 5 = 0 or 5x - 6 = 0
⇒x = -5 or x =

Question 47.
Solve the following quadratic equation for x: (2016D)
Solution:

Question 48.
Solve for x: , x ≠ 1, 2, 3 (2016OD)
Solution:

⇒(x - 1)(x - 3) = 3
⇒x² - 3x - x + 3 - 3 = 0
⇒x² - 4x = 0 ⇒x(x - 4) = 0
⇒x = 0 or x - 4 = 0
∴ x = 0 or x = 4

Question 49.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers. (2016OD)
Solution:
Let three consecutive natural numbers are x, x + 1, x + 2.
According to the question,
(x + 1)² - [(x + 2)² - x²] = 60
⇒x² + 2x + 1 - (x² + 4x + 4 - x²) = 60
⇒x² + 2x + 1 - 4x - 4 - 60 = 0
⇒x² - 2x - 63 = 0
⇒x² - 9x + 7x - 63 = 0
⇒x(x - 9) + 7(x - 9) = 0
⇒(x - 9) (x + 7) = 0
⇒x - 9 = 0 or x + 7 = 0
⇒x = 9 or x = -7
Natural nos. can not be -ve, ∴ x = 9
∴ Numbers are 9, 10, 11.

Question 50.
If the sum of two natural numbers is 8 and their product is 15, find the numbers. (2012OD)
Solution:
Let the numbers be x and (8 - x).
According to the Question,
x(8 - x) = 15
⇒8x - x² = 15
⇒0 = x² - 8x + 15
⇒x² - 5x - 3x + 15 = 0
⇒x(x - 5) - 3(x - 5) = 0
⇒(x - 3)(x - 5) = 0
x - 3 = 0 or x - 5 = 0
x = 3 or x = 5
When x = 3, numbers are 3 and 5.
When x = 5, numbers are 5 and 3.

Quadratic Equations Class 10 Important Questions Long Answer (4 Marks)

Question 51.
Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find the roots. (2014D)
Solution:
(3k + 1)x² + 2(k + 1) + 1 = 0
Here, a = 3k + 1, b = 2(k + 1), c = 1
D = 0 ...[∵ Roots are equal
As b² - 4ac = 0
∴ [2(k + 1)]² - 4(3k + 1)(1) = 0
4(k + 1)² - 4(3k + 1) = 0
4(k² + 2k + 1 - 3k - 1) = 0
(k² - k) = ⇒k(k - 1) = 0
k = 0 or k - 1 = 0
∴ k = 0 or k = 1
Roots are x = ..[As equal roots (Given)
x = ⇒x =
When k = 0, x = = -1
∴ Equal roots are -1 and -1
When k = 1, x =
x=
∴ Equal roots are

Question 52.
Find the value of p for which the quadratic equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 has equal roots. Also find these roots. (2014D)
Solution:
(2p + 1)x² - (7p + 2)x + (7p - 3) = 0
Here, a = 2p + 1, b = -(7p + 2), c = 7p - 3
D = 0 ...[∵ Equal roots As h2 - 4ac = 0
∴ [-(7p + 2)]² - 4(2p + 1)(7p - 3) = 0
⇒(7p + 2)² - 4(14p² - 6p + 7p - 3) = 0
⇒49p² + 28p + 4 - 56p² + 24p - 28p + 12 = 0
⇒-7p² + 24p + 16 = 0
⇒7p² - 24p - 16 = 0 ... [Dividing both sides by -1
⇒7p² - 28p + 4p - 16 = 0
⇒7p(p - 4) + 4(p - 4) = 0
⇒(p - 4) (7p + 4) = 0
⇒p - 4 = 0 or 7p + 4 = 0
⇒p = 4 or p =


∴ Equal roots are 7 and 7.

Question 53.
Find the roots of the equation , x ≠ -4, 7 (2011D)
Solution:

⇒(x + 4)(x - 7) = - 30
⇒x² - 7x + 4x - 28 + 30 = 0
⇒x² - 3x + 2 = 0
⇒x² - x - 2x + 2 = 0
⇒x(x - 1) - 2(x - 1) = 0
⇒(x - 1)(x - 2) = 0
⇒x - 1 = 0 or x - 2 = 0
∴ x = 1 or x = 2

Question 54.
Find the roots of the equation: , x ≠ , 5. (2011OD)
Solution:

Question 55.
Solve for x:
; x ≠ 0, 2, 3 (2013OD)
Solution:


⇒8(x - 3)(x - 2) = x(3x - 8)
⇒8(x² - 3x - 2x +6) = 3x² - 8x
⇒8x² - 24x - 16x + 48 - 3x² + 8x = 0
⇒5x² - 32x + 48 = 0
⇒5x² - 20x - 12x + 48 = 0
⇒5x(x - 4) - 12(x - 4) = 0
⇒(x - 4)(5x - 12) = 0
⇒x - 4 = 0 or 5x - 12 = 0
x = 4 or x =

Question 56.
Solve for x: ; x ≠ 0, (2013OD)
Solution:

⇒5x = (2x + 3 (4 - 3x)
⇒5x = 8x - 6x² + 12 - 9x
⇒5x - 8x + 6x² - 12 + x = 0
⇒6x + 6x - 12 = 0
⇒x² + x - 2 = 0 ...[Dividing by 6
⇒x² + 2x - x - 2 = 0
⇒x(x + 2) - 1(x + 2) = 0
⇒(x - 1)(x + 2) = 0
⇒x - 1 = 0 or x + 2 = 0
∴ x = 1 or x = -2

Question 57.
Solve for x: ; x ≠ 3, 5 (2014OD)
Solution:

⇒4(x² - 8x + 15) = (6x - 24)
⇒4x² - 32x + 60 - 6x + 24 = 0
⇒4x² - 38x + 84 = 0
⇒2x² - 19x + 42 = 0 ...[Dividing by 2
⇒2x² - 12x - 7x + 42 = 0
⇒2x(x -6) - 7(x - 6) = 0
⇒(x - 6) (2x - 7) = 0
⇒x - 6 = 0 or 2x - 7 = 0
∴ x = 6 or x =

Question 58.
Solve for x: (2015D)
Solution:

⇒[3(x - 1) + 4(x + 1)] [4x - 1] = 29(x² - 1)
⇒(3x - 3 + 4x + 4) [4x - 1] = 29(x² - 1)
⇒(7x + 1) (4x - 1) = 29x² - 29
⇒28x² - 3x - 1 = 29x² - 29
⇒x² + 3x - 28 = 0
⇒x² + 7x - 4x - 28 = 0
⇒x(x + 7) - 4(x + 7) = 0
⇒(x + 7) (x - 4) = 0
x = -7 or x = 4
∴ x = -7 and 4

Question 59.
Solve for x: (2015D)
Solution:

⇒5x[4 (x - 2) + 3x + 3) = 46(x + 1) (x - 2)
⇒5x[4x - 8 + 3x + 3) = 46[x² - 1x - 2]
⇒5x (7x - 5) = 46 (x² - x - 2)
⇒35x² - 25x = 46x² - 46x - 92
⇒35x² - 46x² - 25x + 46x + 92 = 0
⇒11x² - 21x - 92 = 0
Here, a = 11, b = -21, c = -92
D = b² - 4ac
= (-21)² - 4 × 11 × (-92)
= 441 + 4048 = 4489

Question 60.
Find x in terms of a, b and c: , x ≠ a, b, c (2016D)
Solution:

⇒(x - c)[ax - ab + bx - ab] = 2c(x - a)(x - b)
⇒(x - c)(ax + bx - 2ab) = 2c(x² - bx - ax + ab)
⇒ax² + bx² - 2abx - acx - bcx + 2abc = 2cx² - 2bcx - 2cax + 2abc
⇒ax² + bx² - 2abx - acx - bcx - 2cx² + 2bcx + 2cax = 0
⇒ax² + bx² - 2cx² - 2abx + bcx + cax = 0
⇒x²(a + b - 2c) + x(-2ab + bc + ca) = 0
⇒x[x (a + b - 2c) + (-2ab + bc + ca)] = 0
⇒x = 0 or x (a + b - 2c) + (-2ab + bc + ca) = 0
⇒x = 0 or x (a + b - 2c) = 2ab - bc - ca = 0
∴ x =

Question 61.
Solve the following for x: (2013D)
Solution:

⇒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 =

Question 62.
A shopkeeper buys some books for 80. If he had bought 4 more books for the same amount, each book would have cost ₹1 less. Find the number of books he bought. (2012D)
Solution:
Let the number of books he bought = x
Increased number of books he had bought = x +4
Total amount = ₹80
According to the problem,

⇒x(x + 4) = 320
⇒x² + 4x - 320 = 0
⇒x² + 20x - 16x - 320 = 0
⇒x(x + 20) - 16(x + 20) = 0
⇒(x + 20) (x - 16) = 0
⇒x + 20 = 0 or x - 16 = 0
⇒x = -20 ... (neglected) or x = 16
∴ Number of books he bought = 16

Question 63.
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. (2013D)
Solution:
Let the side of Large square = x cm
Let the side of small square = y cm
According to the Question,
x² + y² = 400... (i) ...[∵ area of square = (side)²
4x - 4y = 16 ...[∵ Perimeter of square = 4 sides
⇒x - y = 4 ... [Dividing both sides by 4
⇒x = 4 + y ...(ii)
Putting the value of x in equation (i),
(4 + y)² + y2² = 400
⇒y² + 8y + 16 + y² - 400 = 0
⇒2y² + 8y - 384 = 0
⇒y² + 4y - 192 = 0 ... [Dividing both sides by 2
⇒y² + 16y - 12y - 192 = 0
⇒y(y + 16) - 12(y + 16) = 0
⇒(y - 12)(y + 16) = 0
⇒y - 12 = 0 or y + 16 = 0
⇒y = 12 or y = -16 ... [Neglecting negative value
∴ Side of small square = y = 12 cm
and Side of large square = x = 4 + 12 = 16 cm

Question 64.
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 lengths of the sides of the field. (2015OD)
Solution:
Let the length of shorter side be x m.
∴ length of diagonal = (x + 16) m
and length of longer side = (x + 14) m
Using pythagoras theorem,
(l)² + (b)² = (h)²
∴ x² + (x + 14)2² = (x + 16)²
⇒x² + x² + 196 + 28x = x² + 256 + 32x
⇒x² - 4x - 60 = 0
⇒x² - 10x + 6x - 60 = 0
⇒x(x - 10) + 6(x - 10) = 0
⇒(x - 10) (x + 6) = 0
⇒x - 10 = 0 or x + 6 = 0
⇒x = 10 or x = -6 (Reject)
⇒x = 10 m ...[As length cannot be negative
Length of shorter side = x = 10 m
Length of diagonal = (x + 16) m = 26 m
Length of longer side = (x + 14)m = 24m
∴ Length of sides are 10 m and 24 m.

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