# Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations

## Quadratic Equations Exercise Ex. 5(A)

### Solution 1(i)

(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.

### Solution 2(i)

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.

### Solution 2(ii)

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.

### 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)

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.

### Solution 1(iii)

(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.

### Solution 1(iv)

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.

### Solution 1(v)

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.

### Solution 1(vi)

(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.

## Quadratic Equations Exercise Ex. 5(B)

### Solution 1

### Solution 2(i)

### Solution 2(ii)

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

### 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 x^{2} + 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, 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.

### Solution 27

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.

### Solution 28

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

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

Put given value of x in the given equation

_{}

### Solution 29

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### Solution 30

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### Solution 31

### Solution 33

### Solution 34

### Solution 1

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### Solution 2

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### Solution 3

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### Solution 4

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### Solution 5

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### Solution 6

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### Solution 7

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### Solution 8

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### Solution 9

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### Solution 10

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### Solution 11

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### Solution 12

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### Solution 13

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### Solution 14

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### Solution 15

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### Solution 16

### Solution 17

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### Solution 18

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### Solution 19

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### Solution 20

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### 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

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

### Solution 3(iii)

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

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

### Solution 3(v)

## Quadratic Equations Exercise Ex. 5(E)

### Solution 1

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### Solution 2

_{}

### Solution 3

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### Solution 4

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### Solution 5

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### Solution 6

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### Solution 7

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### Solution 8

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### Solution 9

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### Solution 10

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### Solution 11

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### Solution 12

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### Solution 13

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### Solution 14

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### Solution 15

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### Solution 16

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### Solution 17

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### Solution 18

### Solution 19

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### Solution 20

Consider the given equation:

## Quadratic Equations Exercise 5(F)

### Solution 3

Given
quadratic
equation is x^{2} + (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,

x^{2}
+ (3-2a) x - 6a = 0

x^{2}
+ 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}
= 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

### 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 = 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.

### 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,