# Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations

## Quadratic Equations Exercise Ex. 5(A)

### Solution 1(a)

Correct Option: (iv)

4x^{2} – 9 = 0

4x^{2} = 9

### Solution 1(b)

Correct Option: (iii) 3 or –5

(x – 3)(x + 5) = 0

(x – 3) = 0 or (x + 5) = 0

x = 3 or x = –5

### Solution 1(c)

Correct Option: (ii) –3

Substituting x = 4 in equation x^{2} + kx – 4 = 0,

(4)^{2} + k(4) – 4 = 0

16 + 4k – 4 = 0

4k + 12 = 0

4k = –12

k = –3

### Solution 1(d)

Correct Option: (i) –2

Substituting x = 2 in equation 2x^{2} – 3x + k = 0,

2(2)^{2} – 3(2) + k = 0

8 – 6 + k = 0

2 + k = 0

k = –2

### Solution 1(e)

Correct Option: (iv) 0 or 7

x^{2} – 7x = 0

x(x – 7) = 0

x = 0 or x – 7 = 0

x = 0 or x = 7

### Solution 2(i)

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 2(ii)

(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 2(iii)

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 3(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 3(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 4

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 5

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,

## Quadratic Equations Exercise Ex. 5(B)

### Solution 1(a)

Correct Option: (iii) –1 or 7

x^{2} – 6x – 7 = 0

x^{2} – 7x + x – 7 = 0

x(x – 7) + (x – 7) = 0

(x – 7)(x + 1) = 0

x = 7 or x = –1

### Solution 1(b)

Correct Option: (ii) –6 or –2

x(x + 8) + 12 = 0

x^{2} + 8x + 12 = 0

x^{2} + 6x + 2x + 12 = 0

x(x + 6) + 2(x + 6) = 0

(x + 6)(x + 2) = 0

x = –6 or x = –2

### Solution 1(c)

Correct Option: (ii) 2

Substituting x = 2 in equation (p – 3)x^{2} + x + p = 0 is 2,

(p – 3)2^{2} + 2 + p = 0

4p – 12 + 2 + p = 0

5p – 10 = 0

p = 2

### Solution 1(d)

Correct Option: (iii) 2 or ½

x^{2} + 1 = 2.5x

x^{2} – 2.5x + 1 = 0

x^{2} – 2x – 0.5x + 1 = 0

x(x – 2) – 0.5(x – 2) = 0

(x – 2)(x – 0.5) = 0

x = 2 or x = 0.5 = ½

### Solution 1(e)

Correct Option: (i) x ≠ 0

If x = 0, the term in the equation will be undefined.

Hence, for a quadratic equation , x ≠ 0.

### 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

### 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(i)

### Solution 16(ii)

### Solution 17

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

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### Solution 18

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

Put 2x+3 = y

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### Solution 19

or x = -(a + b)

### Solution 20

## Quadratic Equations Exercise Ex. 5(C)

### Solution 1(a)

Correct Option: (ii) 2.0 or 1.0

x^{2} – 3x + 2 = 0

Then, a = 1, b = –3, c = 2

### Solution 1(b)

Correct Option: (i) 5.00 or –1.00

x^{2} – 4x – 5 = 0

Then, a = 1, b = –4, c = –5

### Solution 1(c)

Correct Option: (ii) 9 or –1

x^{2} – 8x – 9 = 0

Then, a = 1, b = –8, c = –9

### Solution 1(d)

Correct Option: (iii) 3.0 or –1.0

x^{2} – 2x – 3 = 0

Then, a = 1, b = –2, c = –3

### Solution 1(e)

Correct Option: (i) 8 or –2

x^{2} – 6x – 16 = 0

Then, a = 1, b = –6, c = –16

### Solution 2

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

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

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

### Solution 4

### Solution 5

### Solution 6

Consider the given equation:

### Solution 7

## Quadratic Equations Exercise Ex. 5(D)

### Solution 1(a)

Correct Option: (i) distinct and real roots

2x^{2} – 3x + 1 = 0

Here, a = 2, b = –3, c = 1

b^{2} – 4ac = (–3)^{2} – 4(2)(1) = 9 – 8 = 1 > 0

Hence, the roots are distinct and real.

### Solution 1(b)

Correct Option: (i) x^{2} – 5x + 6 = 0

For a quadratic equation ax^{2} + bx + c = 0, the roots are real and distinct if b^{2} – 4ac > 0.

For, x^{2} – 5x + 6 = 0

a = 1, b = –5, c = 6

b^{2} – 4ac = (–5)^{2} – 4(1)(6) = 25 – 24 = 1 > 0

Hence, the roots are distinct and real.

### Solution 1(c)

Correct Option: (iv) 4 or –4

For, x^{2} – px + 4 = 0, the roots are equal.

Then, b^{2} – 4ac = 0

Here, a = 1, b = –p, c = 4

b^{2} – 4ac = 0

(–p)^{2} – 4(1)(4) = 0

p^{2} – 16 = 0

p^{2} = 16

p = ±4

### Solution 1(d)

Correct Option: (ii) 2x^{2} – 5x + 9 = 0

For a quadratic equation ax^{2} + bx + c = 0, the roots are imaginary if b^{2} – 4ac < 0.

For, 2x^{2} – 5x + 9 = 0

a = 2, b = –5, c = 9

b^{2} – 4ac = (–5)^{2} – 4(2)(9) = –47 < 0

Hence, the roots are imaginary.

### Solution 1(e)

Correct Option: (i) 7

Substituting x = 1 in equation 3x^{2} – mx + 4 = 0,

3(1)^{2} – m(1) + 4 = 0

3 – m + 4 = 0

m = 7

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

Hence, p = 4 and q = 12 or -12

## Quadratic Equations Exercise Ex. 5(E)

### Solution 1(a)

Correct Option: (ii) ±1 or ±2

x^{4} – 5x^{2} + 4 = 0

Let x^{2} = y

Then,

y^{2} – 5y + 4 = 0

y^{2} – 4y – y + 4 = 0

y(y – 4) – 1(y – 4) = 0

(y – 4)(y – 1) = 0

y = 4 or y = 1

x^{2} = 4 or x^{2} = 1

x = ±2 or x = ±1

### Solution 1(b)

Correct Option: (ii) 10

20x – 50 = 3x^{2} – 15x

3x^{2} – 35x + 50 = 0

3x^{2} – 30x – 5x + 50 = 0

3x(x – 10) – 5(x – 10) = 0

(x – 10)(3x – 5) = 0

x = 10 or

### Solution 1(c)

Correct Option: (iii) x ≠ 3 and x ≠ –5

Here, x – 3 ≠ 0 and x + 5 ≠ 0

x ≠ 3 and x ≠ –5

### Solution 1(d)

Correct Option: (i) –2 and 9 or –9 and 2

Since the product is negative, one integer should be negative.

From the given options, option (i) satisfies the condition.

Hence, the two integers are –2 and 9 or –9 and 2.

### Solution 2

### Solution 3

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### Solution 4 (i)

### Solution 4 (ii)

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

Let x^{2} - 3x = y

Then y^{2 }- 16y - 36 = 0

⇒ y^{2 }- 18y + 2y - 36 = 0

⇒ y(y - 18) + 2(y - 18) = 0

⇒ (y - 18) (y + 2) = 0

If y - 18 = 0 or y + 2 = 0

⇒ x^{2 }- 3x - 18 = 0 or x^{2 }- 3x + 2 =
0

⇒ x^{2 }- 6x + 3x - 18 = 0 or x^{2 }- 2x - x + 2 = 0

⇒ x(x - 6) +3(x - 6) = 0 or x(x - 2) -1(x - 2) = 0

⇒ (x - 6) (x + 3) = 0 or (x - 2) (x - 1) = 0

If x - 6 = 0 or x + 3 = 0 or x - 2 = 0 or x - 1 = 0

then x = 6 or x = -3 or x = 2 or x = 1

### Solution 5

### 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

∴ Given equation reduces to

⇒ 2y^{2} - 3 = 5y

⇒ 2y^{2} - 5y - 3 = 0

⇒ 2y^{2} - 6y + y - 3 =
0

⇒ 2y(y - 3) + 1(y - 3) = 0

⇒ (y - 3)(2y + 1) = 0

⇒ y = 3 and

When, y = 3

⇒ 2x - 1 = 3x + 9

⇒ x = -10

When,

⇒ 4x - 2 = -x - 3

## Quadratic Equations Exercise TEST YOURSELF

### Solution 1(a)

Correct Option: (ii) –4

Consider (3x – 5)(x + 3)

= 3x^{2} – 5x + 9x – 15

= 3x^{2} + 4x – 15

Now, 3x^{2} – kx – 15 = (3x – 5)(x + 3)

That is 3x^{2} – kx – 15 = 3x^{2} + 4x – 15

Therefore, –k = 4 ⇒ k = –4

### Solution 1(b)

Correct Option: (ii) 4

For a quadratic equation ax^{2} + bx + c = 0, the roots are real and equal if b^{2} – 4ac = 0.

For kx^{2} + kx + 1 = 0,

a = k, b = k and c = 1

Now, b^{2} – 4ac = 0

k^{2} – 4(k)(1) = 0

k^{2} – 4k = 0

k(k – 4) = 0

k = 0 or k – 4 = 0 ⇒ k = 4

But for k = 0, the given equation is not satisfied.

Hence, k = 4

(k = 4 is obtained by considering the roots of the equation as real and equal and NOT real and distinct)

### Solution 1(c)

Correct Option: (iii) 5 or –1

x^{2} – 4x = 5

x^{2} – 4x – 5 = 0

x^{2} – 5x + x – 5 = 0

x(x – 5) + 1(x – 5) = 0

(x – 5)(x + 1) = 0

x = 5 or x = –1

### Solution 1(d)

Correct Option: (iv) 0 or 7

x^{2} – 7x = 0

x(x – 7) = 0

x = 0 or x – 7 = 0

x = 0 or x = 7

### Solution 1(e)

Correct Option: (i) 1

Substituting x = 1 in ,

### Solution 2

Given i.e

So, the given quadratic equation becomes

Hence, the values of x are and.

### Solution 3

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

### Solution 4

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### Solution 5

Given quadratic equation is

(i) When the equation has no roots

(ii) When the roots of are

or

### Solution 6

Given quadratic equation is

Using quadratic formula,

⇒ x = a + 1 or x = -a - 2 = -(a + 2)

### Solution 7

Given quadratic equation is

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

and

Hence, .

### Solution 8

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 9

Given quadratic equation is …. (i)

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

Hence, the other root is 2a.

### Solution 10

Given quadratic equation is ….. (i)

Also, given and

and

So, the equation (i) becomes

Hence, the solution of given quadratic equation are and.

### Solution 11

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 12

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### Solution 13

### Solution 14

Consider the given equation:

### Solution 15

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 16

(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 17

Since, -2 is a root of the equation 3x^{2} + 7x + p = 1.

⇒ 3(-2)^{2} + 7(-2) + p = 1

⇒ 12 - 14 + p = 1

⇒ p = 3

The quadratic equation is x^{2} + k(4x + k - 1) + p = 0

i.e. x^{2} + 4kx + k^{2} - k + 3 = 0

Comparing equation x^{2} + 4kx + k^{2} - k + 3 = 0 with ax^{2} + bx + c = 0, we get

a = 1, b = 4k and c = k^{2} - k + 3

Since, the roots are equal.

⇒ b^{2} - 4ac = 0

⇒ (4k)^{2} - 4(k^{2} - k + 3) = 0

⇒ 16k^{2} - 4k^{2} + 4k - 12 = 0

⇒ 12k^{2} + 4k - 12 = 0

⇒ 3k^{2} + k - 3 = 0

By quadratic formula, we have