# ML Aggarwal Solutions for Chapter 7 Quadratic Equations Class 9 Maths ICSE

Here, we are providing the solutions for Chapter 7 Quadratic Equation from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the second chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 7 Quadratic Equation ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on solving equations using factorization.

### Exercise 7

Solve the following (1 to 12) equations:

1. (i) x² – 11x + 30 = 0
(ii) 4x² – 25 = 0

Solution

(i) x² – 11x + 30 = 0

Let us simplify the given equation,

By factorizing, we get

x2 – 5x – 6x + 30 = 0

⇒ x(x – 5) – 6 (x – 5) = 0

⇒ (x – 5) (x – 6) = 0

So,

⇒ (x – 5) = 0 or (x – 6) = 0

⇒ x = 5 or x = 6

∴ Value of x = 5, 6

(ii) 4x² – 25 = 0

Let us simplify the given equation,

4x² = 25

⇒ x2 = 25/4

⇒ x = ±√(25/4)

±5/2

∴ Value of x = +5/2, -5/2

2. (i) 2x² – 5x = 0

(ii) x² – 2x = 48

Solution

(i) 2x² – 5x = 0

Let us simplify the given equation,

x(2x – 5) = 0

So,

⇒ x = 0 or 2x – 5 = 0

⇒ x = 0 or 2x = 5

⇒ x = 0 or x = 5/2

∴ Value of x = 0, 5/2

(ii) x² – 2x = 48

Let us simplify the given equation,

By factorizing, we get

x2 – 2x – 48 = 0

⇒ x2 – 8x+ 6x – 48 = 0

⇒ x(x – 8) + 6 (x – 8) = 0

⇒ (x – 8) (x + 6) = 0

So,

(x – 8) = 0 or (x + 6) = 0

⇒ x = 8 or x = -6

∴ Value of x = 8, -6

3. (i) 6 + x = x²

(ii) 2x² + 3x + 1= 0

Solution

(i) 6 + x = x²

Let us simplify the given equation,

6 + x – x2 = 0

⇒ x2 – x – 6 = 0

By factorizing, we get

x2 – 3x + 2x – 6 = 0

⇒ x(x – 3) + 2 (x – 3) = 0

⇒ (x – 3) (x + 2) = 0

So,

(x – 3) = 0 or (x + 2) = 0

⇒ x = 3 or x = -2

∴ Value of x = 3, -2

(ii) 2x² + 3x + 1= 0

Let us simplify the given equation,

By factorizing, we get

2x2 – 2x – x + 1 = 0

⇒ 2x(x – 1) – 1 (x – 1) = 0

⇒ (x – 1) (2x – 1) = 0

So,

(x – 1) = 0 or (2x – 1) = 0

⇒ x = 1 or 2x = 1

⇒ x = 1 or x = ½

∴ Value of x = 1, ½

4. (i) 3x² = 2x + 8

(ii) 4x² + 15 = 16x

Solution

(i) 3x² = 2x + 8

Let us simplify the given equation,

3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

⇒ 3x(x – 2) + 4 (x – 2) = 0

⇒ (x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

⇒ x = 2 or 3x = -4

⇒ x = 2 or x = -4/3

∴ Value of x = 2 or -4/3

(ii) 4x² + 15 = 16x

Let us simplify the given equation,

4x2 – 16x + 15 = 0

By factorizing, we get

4x2 – 6x – 10x + 15 = 0

⇒ 2x(2x – 3) – 5 (2x – 3) = 0

⇒ (2x – 3) (2x – 5) = 0

So,

(2x – 3) = 0 or (2x – 5) = 0

⇒ 2x = 3 or 2x = 5

⇒ x = 3/2 or x = 5/2

∴ Value of x = 3/2 or 5/2

5. (i) x (2x + 5) = 25

(ii) (x + 3) (x – 3) = 40

Solution

(i) x (2x + 5) = 25

Let us simplify the given equation,

2x2 + 5x – 25 = 0

By factorizing, we get

2x2 + 10x – 5x – 25 = 0

⇒ 2x(x + 5) – 5 (x + 5) = 0

⇒ (x + 5) (2x – 5) = 0

So,

(x + 5) = 0 or (2x – 5) = 0

⇒ x = -5 or 2x = 5

⇒ x = -5 or x = 5/2

∴ Value of x = -5, 5/2

(ii) (x + 3) (x – 3) = 40

Let us simplify the given equation,

x2 – 3x + 3x – 9 = 40

⇒ x2 – 9 – 40 = 0

⇒ x2 – 49 = 0

⇒ x2 = 49

⇒ x = √49

±7

∴ Value of x = 7, -7

6. (i) (2x + 3) (x – 4) = 6

(ii) (3x + 1) (2x + 3) = 3

Solution

(i) (2x + 3) (x – 4) = 6

Let us simplify the given equation,

2x2 – 8x + 3x – 12 – 6 = 0

⇒ 2x2 – 5x – 18 = 0

By factorizing, we get

2x2 – 9x + 4x – 18 = 0

⇒ x(2x – 9) + 2 (2x – 9) = 0

⇒ (2x – 9) (x + 2) = 0

So,

(2x – 9) = 0 or (x + 2) = 0

⇒ 2x = 9 or x = -2

⇒ x = 9/2 or x = -2

∴ Value of x = 9/2, -2

(ii) (3x + 1) (2x + 3) = 3

Let us simplify the given equation,

6x2 + 9x + 2x + 3 – 3 = 0

⇒ 6x2 + 11x = 0

⇒ x(6x + 11) = 0

So,

x = 0 or 6x + 11 = 0

⇒ x = 0 or 6x = -11

⇒ x = 0 or x = -11/6

∴ Value of x = 0, -11/6

7. (i) 4x² + 4x + 1 = 0

(ii) (x – 4)² + 5² = 132

Solution

(i) 4x² + 4x + 1 = 0

Let us simplify the given equation,

By factorizing, we get

4x2 + 2x + 2x + 1 = 0

⇒ 2x(2x + 1) + 1 (2x + 1) = 0

⇒ (2x + 1) (2x + 1) = 0

So,

(2x + 1) = 0 or (2x + 1) = 0

⇒ 2x = -1 or 2x = -1

⇒ x = -1/2 or x = -1/2

∴ Value of x = -1/2, -1/2

(ii) (x – 4)² + 5² = 132

Let us simplify the given equation,

x2 + 16 – 2(x) (4) + 25 – 169 = 0

⇒ x2 – 8x -128 = 0

By factorizing, we get

x2 – 16x + 8x – 128 = 0

⇒ x(x – 16) + 8 (x – 16) = 0

⇒ (x – 16) (x + 8) = 0

So,

(x – 16) = 0 or (x + 8) = 0

⇒ x = 16 or x = -8

∴ Value of x = 16, -8

8. (i) 21x2 = 4 (2x + 1)

(ii) 2/3x2 – 1/3x – 1 = 0

Solution

(i) 21x2 = 4 (2x + 1)

Let us simplify the given equation,

21x2 = 8x + 4

⇒ 21x2 – 8x – 4 = 0

By factorizing, we get

21x2 – 14x + 6x – 4 = 0

⇒ 7x(3x – 2) + 2(3x – 2) = 0

⇒ (3x – 2) (7x + 2) = 0

So,

(3x – 2) = 0 or (7x + 2) = 0

⇒ 3x = 2 or 7x = -2

⇒ x = 2/3 or x = -2/7

∴ Value of x = 2/3 or -2/7

(ii) 2/3x2 – 1/3x – 1 = 0

Let us simplify the given equation,

By taking 3 as LCM and cross multiplying

2x2 – x – 3 = 0

By factorizing, we get

2x2 – 3x + 2x – 3 = 0

⇒ x(2x – 3) + 1 (2x – 3) = 0

⇒ (2x – 3) (x + 1) = 0

So,

(2x – 3) = 0 or (x + 1) = 0

⇒ 2x = 3 or x = -1

⇒ x = 3/2 or x = -1

∴ Value of x = 3/2, -1

9. (i) 6x + 29 = 5/x

(ii) x + 1/x = 2 ½

Solution

(i) 6x + 29 = 5/x

Let us simplify the given equation,

By cross multiplying, we get

6x2 + 29x – 5 = 0

By factorizing, we get

6x2 + 30x – x – 5 = 0

⇒ 6x (x + 5) -1 (x + 5) = 0

⇒ (x + 5) (6x – 1) = 0

So,

(x + 5) = 0 or (6x – 1) = 0

⇒ x = -5 or 6x = 1

⇒ x = -5 or x = 1/6

∴ Value of x = -5, 1/6

(ii) x + 1/x = 2 ½

x + 1/x = 5/2

Let us simplify the given equation,

By taking LCM

x2 + 1 = 5x/2

By cross multiplying,

2x2 + 2 – 5x = 0

⇒ 2x2 – 5x + 2 = 0

By factorizing, we get

2x2 – x – 4x + 2 = 0

⇒ x(2x – 1) – 2 (2x – 1) = 0

⇒ (2x – 1) (x – 2) = 0

So,

(2x – 1) = 0 or (x – 2) = 0

⇒ 2x = 1 or x = 2

⇒ x = ½ or x = 2

∴ Value of x = ½, 2

10. (i) 3x – 8/x = 2

(ii) x/3 + 9/x = 4

Solution

(i) 3x – 8/x = 2

Let us simplify the given equation,

By taking LCM and cross multiplying,

3x2 – 8 = 2x

⇒ 3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

⇒ 3x(x – 2) + 4 (x – 2) = 0

⇒ (x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

⇒ x = 2 or 3x = -4

⇒ x = 2 or x = -4/3

∴ Value of x = 2, -4/3

(ii) x/3 + 9/x = 4

Let us simplify the given equation,

By taking 3x as LCM and cross multiplying

x2 + 27 = 12x

⇒ x2 – 12x + 27 = 0

By factorizing, we get

x2 – 3x – 9x + 27 = 0

⇒ x (x – 3) – 9 (x – 3) = 0

⇒ (x – 3) (x – 9) = 0

So,

(x – 3) = 0 or (x – 9) = 0

⇒ x = 3 or x = 9

∴ Value of x = 3, 9

11. (i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)

(ii) 1/(x + 2) + 1/x = ¾

Solution

(i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)

Let us simplify the given equation,

By cross multiplying,

(x – 1) (3x – 7) = (2x – 5) (x + 1)

⇒ 3x2 – 7x – 3x + 7 = 2x2 + 2x – 5x – 5

⇒ 3x2 – 10x + 7 – 2x2 +3x + 5 = 0

⇒ x2 – 7x + 12 = 0

By factorizing, we get

x2 – 4x – 3x + 12 = 0

⇒ x (x – 4) – 3 (x – 4) = 0

⇒ (x – 4) (x – 3) = 0

So,

(x – 4) = 0 or (x – 3) = 0

⇒ x = 4 or x = 3

∴ Value of x = 4, 3

(ii) 1/(x + 2) + 1/x = ¾

Let us simplify the given equation,

By taking x(x + 2) as LCM

(x+x+2)/x(x + 2) = ¾

By cross multiplying,

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

⇒ 8x + 8= 3x2 + 6x

⇒ 3x2 + 6x – 8x – 8 = 0

⇒ 3x2 – 2x – 8 = 0

By factorizing, we get

3x2 – 6x + 4x – 8 = 0

⇒ 3x(x – 2) + 4 (x – 2) = 0

⇒ (x – 2) (3x + 4) = 0

So,

(x – 2) = 0 or (3x + 4) = 0

⇒ x = 2 or 3x = -4

⇒ x = 2 or x = -4/3

∴ Value of x = 2, -4/3

12. (i) 8/(x + 3) – 3/(2 – x) = 2

(ii) x/(x + 1) + (x + 1)/x = 2 1/6

Solution

(i) 8/(x + 3) – 3/(2 – x) = 2

Let us simplify the given equation,

By taking (x+3)(2-x) as LCM

[8(2-x) – 3(x+3)]/[(x+3)(2-x)] = 2

⇒ [16 – 8x – 3x – 9]/[2x – x2 + 6 – 3x] = 2

⇒ [-11x + 7] = 2(-x2 – x + 6)

⇒ 7 – 11x = -2x2 – 2x + 12

⇒ 2x2 + 2x – 11 x – 12 + 7 = 0

⇒ 2x2 – 9x – 5 = 0

By factorizing, we get

2x2 – 10x + x – 5 = 0

⇒ 2x (x – 5) + 1 (x – 5) = 0

⇒ (x – 5) (2x + 1) = 0

So,

(x – 5) = 0 or (2x + 1) = 0

⇒ x = 5 or 2x= -1

⇒ x = 5 or x = -1/2

∴ Value of x = 5, -1/2

(ii) x/(x + 1) + (x + 1)/x = 2 1/6

⇒ x/(x + 1) + (x + 1)/x = 13/6

Let us simplify the given equation,

By taking x(x+1) as LCM

[x(x) + (x+1) (x+1)]/x(x + 1) = 13/6

⇒ 6[x2 + x2 + x + x + 1] = 13x(x + 1)

⇒ 6[2x2 + 2x + 1] = 13x2 + 13x

⇒ 12x2 + 12x + 6 – 13x2 – 13x = 0

⇒ -x2 – x + 6 = 0

⇒ x2 + x – 6 = 0

By factorizing, we get

x2 + 3x – 2x – 6 = 0

⇒ x (x + 3) – 2 (x + 3) = 0

⇒ (x + 3) (x – 2) = 0

So,

(x + 3) = 0 or (x – 2) = 0

⇒ x = -3 or x = 2

∴ Value of x = -3, 2

### Chapter Test

Solve the following (1 to 3) equations:

1. (i) x (2x + 5) = 3

(ii) 3x2 – 4x – 4 = 0.

Solution

(i) x (2x + 5) = 3

We can write it as

2x2 + 5x – 3 = 0

By further calculation

2x2 + 6x – x – 3 = 0

By taking out the common terms

2x (x + 3) – 1 (x + 3) = 0

So we get

(x + 3) (2x – 1) = 0

Here,

x + 3 = 0 then x = – 3

2x – 1 = 0 then 2x = 1 where x = ½

Therefore, x = – 3, ½.

(ii) 3x2 – 4x – 4 = 0

We can write it as

3x2 – 6x + 2x – 4 = 0

By taking out the common terms

3x (x – 2) + 2 (x – 2) = 0

So we get

(x – 2) (3x + 2) = 0

Here,

x – 2 = 0 then x = 2

3x + 2 = 0 then 3x = -2 where x = -2/3

Therefore, x = 2, -2/3.

2. (i) 4x2 – 2x + ¼ = 0

(ii) 2x2 + 7x + 6 = 0.

Solution

(i) 4x2 – 2x + ¼ = 0

Multiply the equation by 4

16x2 – 8x + 1 = 0

We can write it as

16x2 – 4x – 4x + 1 = 0

Taking out the common terms

4x(4x – 1) – 1 (4x – 1) = 0

So we get

(4x – 1) (4x – 1) = 0

⇒ (4x – 1)2 = 0

Here

4x – 1 = 0

⇒ 4x = 1

By division

x = ¼, ¼

(ii) 2x2 + 7x + 6 = 0

We can write it as

2x2 + 4x + 3x + 6 = 0

By further calculation

2x(x + 2) + 3 (x + 2) = 0

So we get

(x + 2) (2x + 3) = 0

Here

x + 2 = 0 then x = -2

2x + 3 = 0 then 2x = -3 where x = – 3/2

⇒ x = -2, -3/2

3. (i) (x–1)/(x–2) + (x–3)/(x–4) = 3 1/3

(ii) 6/x – 2/(x–1) = 1/(x–2).

Solution

(i) (x–1)/(x–2) + (x–3)/(x–4) = 3 1/3

By taking LCM

[(x–1) (x–4) + (x–2) (x–3)]/(x-2) (x–4) = 10/3

By further calculation

(x2–5x+4+x2–5x+6)/(x2–6x+8) = 10/3

So we get

(2x2–10x+10)/(x2–6x+8) = 10/3

By cross multiplication

10x2 – 60x + 80 = 6x2 – 30x + 30

By further simplification

10x2 – 60x + 80 – 6x2 + 30x – 30 = 0

So we get

4x2 – 30x + 50 = 0

Dividing by 2

2x2 – 15x + 25 = 0

It can be written as

2x2 – 10x – 5x + 25 = 0

Taking out the common terms

2x (x – 5) – 5 (x – 5) = 0

⇒ (x – 5) (2x – 5) = 0

Here

x – 5 = 0 then x = 5

2x – 5 = 0 then 2x = 5 where x = 5/2

Therefore, x = 5, 5/2.

(ii) 6/x – 2/(x – 1) = 1/(x – 2)

Taking LCM

(6x–6–2x)/x(x–1) = 1/(x–2)

By further calculation

(4x – 6)/(x2 – x) = 1/(x – 2)

By cross multiplication

4x2 – 8x – 6x + 12 = x2 – x

So we get

4x2 – 14x + 12 – x2 + x = 0

⇒ 3x2 – 13x + 12 = 0

⇒ 3x2 – 4x – 9x + 12 = 0

Taking out the common terms

x (3x – 4) – 3 (3x – 4) = 0

⇒ (3x – 4) (x – 3) = 0

Here,

3x – 4 = 0 then 3x = 4 where x = 4/3

x – 3 = 0 then x = 3

Therefore, x = 3, 4/3.