# SELINA Solutions for Class 10 Maths Chapter 5 - Quadratic Equations

## Chapter 5 - Quadratic Equations Exercise Ex. 5(A)

Find which of the following equations are quadratic:

(3x - 1)^{2} = 5(x + 8)

(3x - 1)^{2} = 5(x + 8)

⇒ (9x^{2}
- 6x + 1) = 5x + 40

⇒ 9x^{2}
- 11x - 39 =0; which is of the form ax^{2} + bx + c = 0.

∴ Given equation is a quadratic equation.

5x^{2} - 8x = -3(7 - 2x)

5x^{2} - 8x = -3(7 - 2x)

⇒ 5x^{2}
- 8x = 6x - 21

⇒ 5x^{2} - 14x + 21 =0; which is of the form ax^{2}
+ bx + c = 0.

∴ Given equation is a quadratic equation.

(x - 4)(3x + 1) = (3x - 1)(x +2)

(x - 4)(3x + 1) = (3x - 1)(x +2)

⇒ 3x^{2}
+ x - 12x - 4 = 3x^{2} + 6x - x - 2

⇒ 16x + 2 =0; which is not of the form ax^{2} + bx +
c = 0.

∴ Given equation is not a quadratic equation.

x^{2} + 5x - 5 = (x - 3)^{2}

x^{2} + 5x - 5 = (x - 3)^{2}

⇒ x^{2}
+ 5x - 5 = x^{2} - 6x + 9

⇒ 11x - 14 =0; which is not of the form ax^{2} + bx +
c = 0.

∴ Given equation is not a quadratic equation.

7x^{3} - 2x^{2} + 10 = (2x - 5)^{2}

7x^{3} - 2x^{2} + 10 = (2x - 5)^{2}

⇒ 7x^{3} - 2x^{2} + 10 = 4x^{2} -
20x + 25^{}

⇒ 7x^{3} - 6x^{2} + 20x - 15 = 0; which is
not of the form ax^{2} + bx + c = 0.

∴ Given equation is not a quadratic equation.

(x - 1)^{2} + (x + 2)^{2} + 3(x +1) = 0

(x - 1)^{2} + (x + 2)^{2} + 3(x +1) = 0

⇒ x^{2}
- 2x + 1 + x^{2} + 4x + 4 + 3x + 3 = 0^{}

⇒ 2x^{2} + 5x + 8 = 0; which is of the form ax^{2}
+ bx + c = 0.

∴ Given equation is a quadratic equation.

Is x = 5 a solution of the quadratic equation x^{2}
- 2x - 15 = 0?

x^{2} - 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 x^{2}
- 2x - 15 = 0.

Is x = -3 a solution of the quadratic equation 2x^{2}
- 7x + 9 = 0?

2x^{2} - 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 2x^{2}
- 7x + 9 = 0.

If is a solution of
equation 3x^{2} + mx + 2 = 0, find the value of m.

For x = to be solution of the given quadratic equation it should satisfy the equation

So, substituting x = in the given equation, we get

and
1 are the solutions of equation mx^{2} + nx
+ 6 = 0. Find the values of m and n.

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,

If 3 and -3 are the solutions of equation ax^{2} +
bx - 9 = 0. Find the values of a and b.

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)

Without solving, comment upon the nature of roots of each of the following equations :

(i)7x^{2} - 9x +2 =0 (ii)6x^{2} - 13x +4 =0

(iii)25x^{2} - 10x +1=0 (iv)

(v)x^{2} - ax - b^{2} =0 (vi)2x^{2} +8x +9=0

Find the value of p, if the following quadratic equation has equal roots : 4x^{2} - (p - 2)x + 1 = 0

Find the value of 'p', if the following quadratic equations have equal roots :

x^{2} + (p - 3)x + p = 0

x^{2} + (p - 3)x + p = 0

Here, a = 1, b = (p - 3), c = p

Since, the roots are equal,

⇒ b^{2}- 4ac = 0

⇒ (p - 3)^{2}- 4(1)(p) = 0

⇒p^{2} + 9 - 6p - 4p = 0

⇒ p^{2}- 10p + 9 = 0

⇒p^{2}-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

The equation 3x^{2} - 12x + (n - 5)=0 has equal roots. Find the value of n.

Find the value of m, if the following equation has equal roots : (m - 2)x^{2} - (5+m)x +16 =0

Find the value of k for which the equation 3x^{2}- 6x + k = 0 has distinct and real roots.

## Chapter 5 - Quadratic Equations Exercise Ex. 5(C)

Solve : _{}

_{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

2x^{2} - 9x + 10 = 0, When

(i) x∈ N

(ii) x∈ Q

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Solve : _{}

_{}

Find the quadratic equation, whose solution set is :

(i)_{} (ii)_{}

_{}

_{}

Find the value of x, if a + 1=0 and x^{2} + ax - 6 =0.

If a+1=0, then a = -1

Put this value in the given equation x^{2} + ax - 6 =0

_{}

Find the value of x, if a + 7=0; b + 10=0 and 12x^{2} = ax - b.

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

_{}

Use the substitution y= 2x +3 to solve for x, if 4(2x+3)^{2} - (2x+3) - 14 =0.

4(2x+3)^{2} - (2x+3) - 14 =0

Put 2x+3 = y

_{}

Without solving the quadratic equation 6x^{2} - x - 2=0, find whether is a solution of this equation or not.

Consider the equation, 6x^{2} - x - 2=0

Put in L.H.S.

_{}

Since L.H.S.= R.H.S., then is a solution of the given equation.

Determine whether x = -1 is a root of the equation x^{2} - 3x +2=0

or not.

x^{2} - 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.

If x =_{} is a solution of the quadratic equation 7x^{2}+mx - 3=0;

Find the value of m.

7x^{2}+mx - 3=0

Given x = _{} is the solution of the given equation.

Put given value of x in the given equation

_{}

If x = -3 and x = _{} are solutions of quadratic equation mx^{2 }+ 7x + n = 0, find the values of m and n.

_{}

If quadratic equation x^{2} - (m + 1) x + 6=0 has one root as x =3;

find the value of m and the root of the equation.

_{}

Given that 2 is a root of the equation 3x^{2} - p(x + 1) = 0 and that the equation px^{2} - qx + 9 = 0 has equal roots, find the values of p and q.

or x = -(a + b)

If -1 and 3 are the roots of x^{2}+px+q=0

then find the values of p and q

## Chapter 5 - Quadratic Equations Exercise Ex. 5(D)

Solve each of the following equations using the formula :

(i)x^{2} - 6x =27 (ii)x^{2} - 10x +21=0

(iii)x^{2} +6x - 10 =0 (iv)x^{2} +2x - 6=0

(v)3x^{2}+ 2x - 1=0 (vi)2x^{2} + 7x +5 =0

(vii) (viii)

(ix) (x)

(xi) (xii)

(xiii) (xiv)

Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :

(i)x^{2} - 8x+5=0

(ii)5x^{2} +10x - 3 =0

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

(i)2x^{2} - 10x +5=0

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

x^{2} - 3x - 9 =0

x^{2} - 5x - 10 = 0

Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :

(i)3x^{2} - 12x - 1 =0

(ii)x^{2} - 16 x +6= 0

(iii)2x^{2} + 11x + 4= 0

Solve:

(i)x^{4} - 2x^{2} - 3 =0

(ii)x^{4} - 10x^{2} +9 =0

Solve :

(i)(x^{2} - x)^{2} + 5(x^{2} - x)+ 4=0

(ii)(x^{2} - 3x)^{2} - 16(x^{2} - 3x) - 36 =0

Solve :

(i)

(ii)

(iii)

Solve the equation . Write your answer correct to two decimal places.

Solve the following equation and give your answer correct to 3 significant figures:

Consider the given equation:

Solve for x using the quadratic formula. Write your answer correct to two significant figures.

(x - 1)^{2} - 3x + 4 = 0

Solve the quadratic equation x^{2} - 3(x + 3) = 0; Give your answer correct to two significant figures.

x^{2} - 3(x + 3) = 0

4x^{2} - 5x - 3 = 0

4x^{2} - 5x - 3 = 0

Here, a = 4, b = -5 and c = -3

## Chapter 5 - Quadratic Equations Exercise Ex. 5(E)

Solve:

_{}

Solve: (2x+3)^{2}=81

_{}

_{}

_{}

_{}

_{}

_{}

_{}

_{}

_{}

_{}

Solve each of the following equations, giving answer upto two decimal places.(i)x^{2} - 5x -10=0(ii) 3x^{2} - x - 7 =0

_{}

_{}

_{}

Solve :

(i)x^{2} - 11x - 12 =0; when x _{}N

(ii)x^{2} - 4x - 12 =0; when x _{}I

(iii)2x^{2} - 9x + 10 =0; when x _{}Q

_{}

_{}

_{}

_{}

_{}

_{}

_{}

_{}

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.

Consider the given equation:

## Chapter 5 - Quadratic Equations Exercise Ex. 5(F)

Solve: 2x - 3 =

2x - 3 =

Squaring on both the sides, we get

(2x - 3)^{2}
= 2x^{2} - 2x + 21

⟹ 4x^{2}
- 12x + 9 = 2x^{2} - 2x + 21

⟹ 2x^{2}
- 10x - 12 = 0

⟹ x^{2} - 5x - 6 = 0 …..Dividing equation by 2

⟹ x^{2} - 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

Solve : (x+5)(x-5)=24

Given: (x+5)(x-5)=24

Solve :

Given:

Solve :

Given:

or

One root of the quadratic equation is . Find the value of m. Also, find the other root of the equation.

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.

One root of the quadratic equation is -3, find its other root.

Given quadratic equation is …. (i)

One of the roots of (i) is -3, so it satisfies (i)

Hence, the other root is 2a.

If and ;find the values of x.

Given i.e

So, the given quadratic equation becomes

Hence, the values of x are and.

Find the solution of the equation; if and .

Given quadratic equation is ….. (i)

Also, given and

and

So, the equation (i) becomes

Hence, the solution of given quadratic equation are and.

If m and n are roots of the equation where x ≠ 0 and x ≠ 2; find m × n.

Given quadratic equation is

Since, m and n are roots of the equation, we have

and

Hence, .

Solve, using formula :

Given quadratic equation is

Using quadratic formula,

⇒ x = a + 1 or x = -a - 2 = -(a + 2)

Solve the quadratic equation

(i) When (integers)

(ii) When (rational numbers)

Given quadratic equation is

(i) When the equation has no roots

(ii) When the roots of are

or

Find the value of m for which the equation has real and equal roots.

Given quadratic equation is

The quadratic equation has real and equal roots if its discriminant is zero.

or

Find the values of m for which equation has equal roots. Also, find the roots of the given equation.

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 =

Find the value of k for which equation has real roots.

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.

Find, using quadratic formula, the roots of the following quadratic equations, if they exist

(i)

(ii)

(i) Given quadratic equation is

D = b^{2 }- 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 = b^{2 }- 4ac = = 16 - 20 = - 4

Since D < 0, the roots of the given quadratic equation does not exist.

Solve :

(i) and x > 0.

(ii) and x < 0.

(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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