SELINA Solutions for Class 10 Maths Chapter 5 - Quadratic Equations
Aim for top marks in the prelim and board exams with the Selina Solutions for ICSE Class 10 Mathematics Chapter 5 Quadratic Equations. Improve your analytical abilities to prove whether the given equation is a quadratic equation or not using our textbook solutions. Also, learn the concept of two equal roots and two distinct real roots through Maths practice.
Revise the quadratic formula and the factorisation method to prove quadratic equations with the support of our ICSE Class 10 Maths textbook solutions. To revisit the problem-solving methods, you can watch our concept videos or go through our well-written revision notes.
Chapter 5 - Quadratic Equations Exercise Ex. 5(A)
Question 1(iv)
x2 + 5x - 5 = (x - 3)2
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.
Question 1(v)
7x3 - 2x2 + 10 = (2x - 5)2
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.
Question 1(vi)
(x - 1)2 + (x + 2)2 + 3(x +1) = 0
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.
Question 1(i)
Find which of the following equations are quadratic:
(3x - 1)2 = 5(x + 8)
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.
Question 1(ii)
5x2 - 8x = -3(7 - 2x)
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.
Question 1(iii)
(x - 4)(3x + 1) = (3x - 1)(x +2)
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.
Question 2(i)
Is x = 5 a solution of the quadratic equation x2 - 2x - 15 = 0?
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.
Question 2(ii)
Is x = -3 a solution of the quadratic equation 2x2 - 7x + 9 = 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.
Question 3
If 
is a solution of
equation 3x2 + mx + 2 = 0, find the value of m.
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
Question 4
and
1 are the solutions of equation mx2 + nx
+ 6 = 0. Find the values of m and n.
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,
Question 5
If 3 and -3 are the solutions of equation ax2 + bx - 9 = 0. Find the values of a and b.
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,
Chapter 5 - Quadratic Equations Exercise Ex. 5(B)
Question 1
Without solving, comment upon the nature of roots of each of the following equations :
(i)7x2 - 9x +2 =0 (ii)6x2 - 13x +4 =0
(iii)25x2 - 10x +1=0 (iv)
(v)x2 - ax - b2 =0 (vi)2x2 +8x +9=0
Solution 1
Question 2(i)
Find the value of p, if the following quadratic equation has equal roots : 4x2 - (p - 2)x + 1 = 0
Solution 2(i)
Question 2(ii)
Find the value of 'p', if the following quadratic equations have equal roots :
x2 + (p - 3)x + p = 0
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
Question 3
The equation 3x2 - 12x + (n - 5)=0 has equal roots. Find the value of n.
Solution 3
Question 4
Find the value of m, if the following equation has equal roots : (m - 2)x2 - (5+m)x +16 =0
Solution 4
Question 5
Find the value of k for which the equation 3x2- 6x + k = 0 has distinct and real roots.
Solution 5
Chapter 5 - Quadratic Equations Exercise Ex. 5(C)
Question 1
Solve : 
Solution 1
Question 2
Solve : 
Solution 2
Question 3
Solve : 
Solution 3
Question 4
Solve : 
Solution 4
Question 5
Solve : 
Solution 5
Question 6
Solve : 
Solution 6
Question 7
Solve : 
Solution 7
Question 8
Solve : 
Solution 8
Question 9
Solve : 
Solution 9
Question 10
Solve : 
Solution 10
Question 11
Solve : 
Solution 11
Question 12
Solve : 
Solution 12
Question 13
Solve : 
Solution 13
Question 14
Solve : 
Solution 14
Question 15
Solve : 
Solution 15
Question 16
2x2 - 9x + 10 = 0, When
(i) x∈ N
(ii) x∈ Q
Solution 16
Question 17
Solve : 
Solution 17
Question 18
Solve : 
Solution 18
Question 19
Solve : 
Solution 19
Question 20
Solve : 
Solution 20
Question 21
Find the quadratic equation, whose solution set is :
(i)
(ii)
Solution 21
Question 22(i)
Solution 22(i)
Question 22(ii)
Solution 22(ii)
Question 23
Find the value of x, if a + 1=0 and x2 + ax - 6 =0.
Solution 23
If a+1=0, then a = -1
Put this value in the given equation x2 + ax - 6 =0
Question 24
Find the value of x, if a + 7=0; b + 10=0 and 12x2 = ax - b.
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
Question 25
Use the substitution y= 2x +3 to solve for x, if 4(2x+3)2 - (2x+3) - 14 =0.
Solution 25
4(2x+3)2 - (2x+3) - 14 =0
Put 2x+3 = y
Question 26
Without solving the quadratic equation 6x2 - x - 2=0, find whether 
is a solution of this equation or not.
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.
Question 27
Determine whether x = -1 is a root of the equation x2 - 3x +2=0
or not.
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.
Question 28
If x =
is a solution of the quadratic equation 7x2+mx - 3=0;
Find the value of m.
Solution 28
7x2+mx - 3=0
Given x = 
is the solution of the given equation.
Put given value of x in the given equation
Question 29
If x = -3 and x = 
are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.
Solution 29
Question 30
If quadratic equation x2 - (m + 1) x + 6=0 has one root as x =3;
find the value of m and the root of the equation.
Solution 30
Question 31
Given that 2 is a root of the equation 3x2 - p(x + 1) = 0 and that the equation px2 - qx + 9 = 0 has equal roots, find the values of p and q.
Solution 31
Question 32
Solution 32
or x = -(a + b)
Question 33
Solution 33
Question 34
If -1 and 3 are the roots of x2+px+q=0
then find the values of p and q
Solution 34
Chapter 5 - Quadratic Equations Exercise Ex. 5(D)
Question 1
Solve each of the following equations using the formula :
(i)x2 - 6x =27 (ii)x2 - 10x +21=0
(iii)x2 +6x - 10 =0 (iv)x2 +2x - 6=0
(v)3x2+ 2x - 1=0 (vi)2x2 + 7x +5 =0
(vii)
(viii) 
(ix) 
(x) 
(xi) 
(xii) 
(xiii)
(xiv)
Solution 1
Question 2
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :
(i)x2 - 8x+5=0
(ii)5x2 +10x - 3 =0
Solution 2
Question 3(iv)
x2 - 5x - 10 = 0
Solution 3(iv)
Question 3(iii)
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
x2 - 3x - 9 =0
Solution 3(iii)
Question 3(ii)
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
Solution 3(ii)
Question 3(i)
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
(i)2x2 - 10x +5=0
Solution 3(i)
Question 4
Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :
(i)3x2 - 12x - 1 =0
(ii)x2 - 16 x +6= 0
(iii)2x2 + 11x + 4= 0
Solution 4
Question 5
Solve:
(i)x4 - 2x2 - 3 =0
(ii)x4 - 10x2 +9 =0
Solution 5
Question 6
Solve :
(i)(x2 - x)2 + 5(x2 - x)+ 4=0
(ii)(x2 - 3x)2 - 16(x2 - 3x) - 36 =0
Solution 6
Question 7
Solve :
(i)
(ii)
(iii)
Solution 7
Question 8
Solve the equation 
. Write your answer correct to two decimal places.
Solution 8
Question 9
Solve the following equation and give your answer correct to 3 significant figures:
Solution 9
Consider the given equation:
Question 10
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x - 1)2 - 3x + 4 = 0
Solution 10
Question 11
Solve the quadratic equation x2 - 3(x + 3) = 0; Give your answer correct to two significant figures.
Solution 11
x2 - 3(x + 3) = 0
Chapter 5 - Quadratic Equations Exercise Ex. 5(E)
Question 1
Solve:
Solution 1
Question 2
Solve: (2x+3)2=81
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solve each of the following equations, giving answer upto two decimal places.(i)x2 - 5x -10=0(ii) 3x2 - x - 7 =0
Solution 12
Question 13
Solution 13
Question 14
Solve :
(i)x2 - 11x - 12 =0; when x 
N
(ii)x2 - 4x - 12 =0; when x I
(iii)2x2 - 9x + 10 =0; when x Q
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
Solution 20
Consider the given equation:
Chapter 5 - Quadratic Equations Exercise Ex. 5(F)
Question 1(i)
Solve : (x+5)(x-5)=24
Solution 1(i)
Given: (x+5)(x-5)=24
Question 1(ii)
Solve : 
Solution 1(ii)
Given:
Question 1(iii)
Solve : 
Solution 1(iii)
Given:
or 
Question 2
One root of the quadratic equation 
is
. Find the value of m. Also, find the other root of the
equation.
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
.
Question 3
One root of the quadratic equation 
is -3, find its
other root.
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.
Question 4
If 
and 
;find the values of x.
Solution 4
Given i.e
So, the given quadratic equation becomes
Hence, the values of x are 
and
.
Question 5
Find the solution of the equation
; if 
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.
Question 6
If m and n are roots of the equation 
where x ≠ 0 and x ≠ 2; find m × n.
Solution 6
Given quadratic equation is 
Since, m and n are roots of the equation, we have
and 
Hence, 
.
Question 7
Solve, using formula :
Solution 7
Given quadratic equation is 
Using quadratic formula,
⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Question 8
Solve the quadratic equation 
(i) When
(integers)
(ii) When 
(rational numbers)
Solution 8
Given quadratic equation is
(i) When 
the equation has no roots
(ii) When 
the roots of are
or 
Question 9
Find the value of m for which the equation 
has real and equal
roots.
Solution 9
Given quadratic equation is
The quadratic equation has real and equal roots if its discriminant is zero.
or 
Question 10
Find the values of m for which equation 
has equal roots.
Also, find the roots of the given equation.
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 = 
Question 11
Find the value of k for which equation 
has real roots.
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
.
Question 12
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i) 
(ii) 
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.
Question 13
Solve :
(i) 
and
x > 0.
(ii) 
and x < 0.
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, 
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