Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations
Quadratic Equations Exercise Ex. 5(A)
Solution 1(i)
(3x - 1)2 = 5(x + 8)
⇒ (9x2 - 6x + 1) = 5x + 40
⇒ 9x2 - 11x - 39 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
Solution 2(i)
x2 - 2x - 15 = 0
For x = 5 to be solution of the given quadratic equation it should satisfy the equation.
So, substituting x = 5 in the given equation, we get
L.H.S = (5)2 - 2(5) - 15
= 25 - 10 - 15
= 0
= R.H.S
Hence, x = 5 is a solution of the quadratic equation x2 - 2x - 15 = 0.
Solution 2(ii)
2x2 - 7x + 9 = 0
For x = -3 to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = 5 in the given equation, we get
L.H.S=2(-3)2 - 7(-3) + 9
= 18 + 21 + 9
= 48
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0.
Solution 3
For x = 
to be solution of
the given quadratic equation it should satisfy the equation
So, substituting x = in the given
equation, we get
Solution 4
For x = 
and x = 1 to be
solutions of the given quadratic equation it should satisfy the equation
So, substituting x = and x = 1 in the
given equation, we get
Solving equations (1) and (2) simultaneously,
Solution 5
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 3 and x = -3 in the given equation, we get
Solving equations (1) and (2) simultaneously,
Solution 1(ii)
5x2 - 8x = -3(7 - 2x)
⇒ 5x2 - 8x = 6x - 21
⇒ 5x2 - 14x + 21 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
Solution 1(iii)
(x - 4)(3x + 1) = (3x - 1)(x +2)
⇒ 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2
⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Solution 1(iv)
x2 + 5x - 5 = (x - 3)2
⇒ x2 + 5x - 5 = x2 - 6x + 9
⇒ 11x - 14 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Solution 1(v)
7x3 - 2x2 + 10 = (2x - 5)2
⇒ 7x3 - 2x2 + 10 = 4x2 - 20x + 25
⇒ 7x3 - 6x2 + 20x - 15 = 0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Solution 1(vi)
(x - 1)2 + (x + 2)2 + 3(x +1) = 0
⇒ x2 - 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0
⇒ 2x2 + 5x + 8 = 0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
Quadratic Equations Exercise Ex. 5(B)
Solution 1
Solution 2(i)
Solution 2(ii)
x2 + (p - 3)x + p = 0
Here, a = 1, b = (p - 3), c = p
Since, the roots are equal,
⇒ b2- 4ac = 0
⇒ (p - 3)2- 4(1)(p) = 0
⇒p2 + 9 - 6p - 4p = 0
⇒ p2- 10p + 9 = 0
⇒p2-9p - p + 9 = 0
⇒p(p - 9) - 1(p - 9) = 0
⇒ (p -9)(p - 1) = 0
⇒ p - 9 = 0 or p - 1 = 0
⇒ p = 9 or p = 1
Solution 3
Solution 4
Solution 5
Quadratic Equations Exercise Ex. 5(C)
Solution 21
Solution 22(i)
Solution 22(ii)
Solution 23
If a+1=0, then a = -1
Put this value in the given equation x2 + ax - 6 =0
Solution 24
If a + 7 =0, then a = -7
and b + 10 =0, then b = - 10
Put these values of a and b in the given equation
Solution 25
4(2x+3)2 - (2x+3) - 14 =0
Put 2x+3 = y
Solution 26
Consider the equation, 6x2 - x - 2=0
Put 
in L.H.S.
Since L.H.S.= R.H.S., then is a solution of the given equation.
Solution 27
x2 - 3x +2=0
Put x = -1 in L.H.S.
L.H.S. = (-1)2 - 3(-1) +2
= 1 +3 +2=6 
R.H.S.
Then x = -1 is not the solution of the given equation.
Solution 28
7x2+mx - 3=0
Given x = 
is the solution of the given equation.
Put given value of x in the given equation
Solution 29
Solution 30
Solution 31
Solution 33
Solution 34
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 32
or x = -(a + b)
Quadratic Equations Exercise Ex. 5(D)
Solution 1
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iv)
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Consider the given equation:
Solution 10
Solution 11
x2 - 3(x + 3) = 0
Solution 3(iii)
4x2 - 5x - 3 = 0
Here, a = 4, b = -5 and c = -3
Solution 3(v)
Quadratic Equations Exercise Ex. 5(E)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Consider the given equation:
Quadratic Equations Exercise 5(F)
Solution 3
Given quadratic equation is x2 + (3-2a) x - 6a = 0
Now substitute x = -3
So, (-3)2 + (3-2a)(-3) - 6a = 0
0=0
Hence x = -3 satisfies the equation, so -3 is a root.
Now,
x2 + (3-2a) x - 6a = 0
x2 + 3x-2ax - 6a = 0
x(x+3)-2a(x+3)=0
(x-2a)(x+3)=0
x=-3, 2a
Hence, the other root is 2a.
Quadratic Equations Exercise Ex. 5(F)
Solution 1(iv)
2x - 3 = 
Squaring on both the sides, we get
(2x - 3)2 = 2x2 - 2x + 21
⟹ 4x2 - 12x + 9 = 2x2 - 2x + 21
⟹ 2x2 - 10x - 12 = 0
⟹ x2 - 5x - 6 = 0 …..Dividing equation by 2
⟹ x2 - 6x + x - 6 = 0
⟹ x(x - 6) + 1(x - 6) = 0
⟹ (x - 6)(x + 1) = 0
⟹ (x - 6) = 0 or (x + 1) = 0
⟹ x = 6 or x = -1
Solution 1(i)
Given: (x+5)(x-5)=24
Solution 1(ii)
Given: 
Solution 1(iii)
Given: 
or 
Solution 2
Given quadratic equation is 
…. (i)
One of the roots of (i) is
, so it satisfies (i)
So, the equation (i)
becomes 
Hence, the other root is
.
Solution 3
Given quadratic equation is 
…. (i)
One of the roots of (i) is -3, so it satisfies (i)
Hence, the other root is 2a.
Solution 4
Given 
i.e
So, the given quadratic equation becomes
Hence, the values of x are 
and
.
Solution 5
Given quadratic equation is 
….. (i)
Also, given 
and 
and 
So, the equation (i) becomes
Hence, the solution of given quadratic equation are and.
Solution 6
Given quadratic equation is 
Since, m and n are roots of the equation, we have
and 
Hence, 
.
Solution 7
Given quadratic equation is 
Using quadratic formula,
⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Solution 8
Given quadratic equation is 
(i) When 
the equation has no roots
(ii) When 
the roots of are
or 
Solution 9
Given quadratic equation is 
The quadratic equation has real and equal roots if its discriminant is zero.
or 
Solution 10
Given quadratic equation is 
…. (i)
The quadratic equation has equal roots if its discriminant is zero
When 
, equation (i) becomes
When 
, equation (i) becomes
∴ x = 
Solution 11
Given quadratic equation is 
…. (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
Hence, the given quadratic equation has real roots for
.
Solution 12
(i) Given quadratic
equation is 
D = b2 - 4ac =
= 25 - 24 = 1
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have
or 
(ii) Given quadratic equation is 
D = b2 - 4ac =
= 16 - 20 = - 4
Since D < 0, the roots of the given quadratic equation does not exist.
Solution 13
(i) Given quadratic
equation is 
or 
But as x > 0, so x can't be negative.
Hence, x = 6.
(ii) Given quadratic equation is 
or 
But as x < 0, so x can't be positive.
Hence, 