SELINA Solutions for Class 10 Maths Chapter 5 - Quadratic Equations
Chapter 5 - Quadratic Equations Exercise Ex. 5(A)
Question 1
x2 + 5x - 5 = (x - 3)2
Solution 1
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 2
7x3 - 2x2 + 10 = (2x - 5)2
Solution 2
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 3
(x - 1)2 + (x + 2)2 + 3(x +1) = 0
Solution 3
(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 4
Find which of the following equations are quadratic:
(3x - 1)2 = 5(x + 8)
Solution 4
(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 5
5x2 - 8x = -3(7 - 2x)
Solution 5
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 6
(x - 4)(3x + 1) = (3x - 1)(x +2)
Solution 6
(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 7
Is x = 5 a solution of the quadratic equation x2 - 2x - 15 = 0?
Solution 7
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 8
Is x = -3 a solution of the quadratic equation 2x2 - 7x + 9 = 0?
Solution 8
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 9
If 
is a solution of
equation 3x2 + mx + 2 = 0, find the value of m.
Solution 9
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 10
and
1 are the solutions of equation mx2 + nx
+ 6 = 0. Find the values of m and n.
Solution 10
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 11
If 3 and -3 are the solutions of equation ax2 + bx - 9 = 0. Find the values of a and b.
Solution 11
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
Find the value of p, if the following quadratic equation has equal roots : 4x2 - (p - 2)x + 1 = 0
Solution 2
Question 3
Find the value of 'p', if the following quadratic equations have equal roots :
x2 + (p - 3)x + p = 0
Solution 3
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 4
The equation 3x2 - 12x + (n - 5)=0 has equal roots. Find the value of n.
Solution 4
Question 5
Find the value of m, if the following equation has equal roots : (m - 2)x2 - (5+m)x +16 =0
Solution 5
Question 6
Find the value of k for which the equation 3x2- 6x + k = 0 has distinct and real roots.
Solution 6
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
Solution 22
Question 23
Solution 23
Question 24
Find the value of x, if a + 1=0 and x2 + ax - 6 =0.
Solution 24
If a+1=0, then a = -1
Put this value in the given equation x2 + ax - 6 =0
Question 25
Find the value of x, if a + 7=0; b + 10=0 and 12x2 = ax - b.
Solution 25
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 26
Use the substitution y= 2x +3 to solve for x, if 4(2x+3)2 - (2x+3) - 14 =0.
Solution 26
4(2x+3)2 - (2x+3) - 14 =0
Put 2x+3 = y
Question 27
Without solving the quadratic equation 6x2 - x - 2=0, find whether 
is a solution of this equation or not.
Solution 27
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 28
Determine whether x = -1 is a root of the equation x2 - 3x +2=0
or not.
Solution 28
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 29
If x =
is a solution of the quadratic equation 7x2+mx - 3=0;
Find the value of m.
Solution 29
7x2+mx - 3=0
Given x = 
is the solution of the given equation.
Put given value of x in the given equation
Question 30
If x = -3 and x = 
are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.
Solution 30
Question 31
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 31
Question 32
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 32
Question 33
Solution 33
or x = -(a + b)
Question 34
Solution 34
Question 35
If -1 and 3 are the roots of x2+px+q=0
then find the values of p and q
Solution 35
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
x2 - 5x - 10 = 0
Solution 3
Question 4
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 4
Question 5
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
Solution 5
Question 6
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 6
Question 7
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 7
Question 8
Solve:
(i)x4 - 2x2 - 3 =0
(ii)x4 - 10x2 +9 =0
Solution 8
Question 9
Solve :
(i)(x2 - x)2 + 5(x2 - x)+ 4=0
(ii)(x2 - 3x)2 - 16(x2 - 3x) - 36 =0
Solution 9
Question 10
Solve :
(i)
(ii)
(iii)
Solution 10
Question 11
Solve the equation 
. Write your answer correct to two decimal places.
Solution 11
Question 12
Solve the following equation and give your answer correct to 3 significant figures:
Solution 12
Consider the given equation:
Question 13
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x - 1)2 - 3x + 4 = 0
Solution 13
Question 14
Solve the quadratic equation x2 - 3(x + 3) = 0; Give your answer correct to two significant figures.
Solution 14
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
Solve : (x+5)(x-5)=24
Solution 1
Given: (x+5)(x-5)=24
Question 2
Solve : 
Solution 2
Given:
Question 3
Solve : 
Solution 3
Given:
or 
Question 4
One root of the quadratic equation 
is
. Find the value of m. Also, find the other root of the
equation.
Solution 4
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 5
One root of the quadratic equation 
is -3, find its
other root.
Solution 5
Given quadratic equation is …. (i)
One of the roots of (i) is -3, so it satisfies (i)
Hence, the other root is 2a.
Question 6
If 
and 
;find the values of x.
Solution 6
Given i.e
So, the given quadratic equation becomes
Hence, the values of x are 
and
.
Question 7
Find the solution of the equation
; if 
and 
.
Solution 7
Given quadratic equation is ….. (i)
Also, given and
and 
So, the equation (i) becomes
Hence, the solution of given quadratic equation are and.
Question 8
If m and n are roots of the equation 
where x ≠ 0 and x ≠ 2; find m × n.
Solution 8
Given quadratic equation is 
Since, m and n are roots of the equation, we have
and 
Hence, 
.
Question 9
Solve, using formula :
Solution 9
Given quadratic equation is 
Using quadratic formula,
⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Question 10
Solve the quadratic equation 
(i) When
(integers)
(ii) When 
(rational numbers)
Solution 10
Given quadratic equation is
(i) When 
the equation has no roots
(ii) When 
the roots of are
or 
Question 11
Find the value of m for which the equation 
has real and equal
roots.
Solution 11
Given quadratic equation is
The quadratic equation has real and equal roots if its discriminant is zero.
or 
Question 12
Find the values of m for which equation 
has equal roots.
Also, find the roots of the given equation.
Solution 12
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 13
Find the value of k for which equation 
has real roots.
Solution 13
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 14
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i) 
(ii) 
Solution 14
(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 15
Solve :
(i) 
and
x > 0.
(ii) 
and x < 0.
Solution 15
(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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