Important Questions for Class 10 Maths Chapter 4 Quadratic Equations
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 =
Question 3.
If x = –
Solution:
The given quadratic equation can be written as, 3x2 + 2kx – 3 = 0
Question 4.
If the quadratic equation px\frac{1}{2} – 2
Solution:
The given quadratic equation can be written as px\frac{1}{2} – 2
Here a = p, b = – 2
For equal roots, D = 0
D = b2 – 4ac – 0 …[∵ Equal roots
0 = (-2
0 = 4 × 5p2 – 60p
0 = 20p2 – 60p => 20p2 = 60p
p =
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
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 =
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
Solution:
4
4
4x(
(
Question 13.
Solve the following for x:
Solution:
⇒
x(
(
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: 4x2– 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 =
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 =
Solution:
We have, ax2 + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
Now, α =
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:
Solution:
⇒
⇒ (
⇒ 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)
5 + 8 = 13 ⇒ 13 = 13 …[True
∴ x = 8 is the solution.
Checking, When x = 20 in (i),
7 + 20 ≠ 13 …[False
∴ x = 20 is not a solution.
Therefore, x = 8 is the only solution.
Question 24.
Solve for x:
Solution:
⇒
⇒ (
⇒ 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),
7 – 7 = 0 … [True
Checking: When x =
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
Solution:
We have, 2
Here, a = 2
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 =
Question 27.
Solve for x: 3x2} – 2
Solution:
3x2} – 2
⇒ 3x2 –
⇒
⇒ (
⇒
∴ 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 =
∴ 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
Solution:
We have, kx(x – 2
kx2 – 2
Here a = k, b= -2
D = 0 …[∵ Roots are equal
As b2 – 4ac = 0
∴ (-2
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 =
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:
Solution:
The given quadratic equation can be written as
Here, a =
D = b2 – 4ac
Question 40.
Solve for x: 2x2 + 6
Solution:
Given equation can be written as
2(x2 + 3
x2 + 3
Here, a = 1, b = 3
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
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
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
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:
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=
But, x ≠
∴ x= -1 is the only solution.
Question 46.
Solve for x: