ML Aggarwal Solutions for Class 9 Maths Chapter 7 – Quadratic Equations are provided here to help students understand all the concepts clearly and develop a strong command over the subject. This chapter mainly deals with problems based on quadratic equations. Sometimes, students face difficulty in understanding the concepts during class hours, for those students, the experts at BYJUâ€™S have designed the solutions based on the studentsâ€™ grasping abilities. Students can solve both chapter-wise and exercise wise problems to increase their confidence level before appearing for the board exam. To boost interest among students in this chapter, ML Aggarwal Solutions for Class 9 Chapter 7 quadratic equations pdf links are given here for easy access. Students can download the pdfs easily and can use it for future reference as well.

Chapter 7 – Quadratic Equations contains one exercise and the

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EXERCISE 7

**Solve the following (1 to 12) equations: **

**1. (i) xÂ² â€“ 11x + 30 = 0(ii) 4xÂ² – 25 = 0**

**Solution:**

**(i) **xÂ² â€“ 11x + 30 = 0

Let us simplify the given equation,

By factorizing, we get

x^{2} â€“ 5x â€“ 6x + 30 = 0

x(x – 5) â€“ 6 (x – 5) = 0

(x â€“ 5) (x â€“ 6) = 0

So,

(x â€“ 5) = 0 or (x â€“ 6) = 0

x = 5 or x = 6

âˆ´ Value of x = 5, 6

**(ii) **4xÂ² – 25 = 0

Let us simplify the given equation,

4xÂ² = 25

x^{2} = 25/4

x = **Â±** âˆš(25/4)

= **Â±**5/2

âˆ´ Value of x = +5/2, -5/2

**2. (i) 2xÂ² â€“ 5x = 0**

**(ii) xÂ² â€“ 2x = 48**

**Solution:**

**(i) **2xÂ² â€“ 5x = 0

Let us simplify the given equation,

x(2x – 5) = 0

so,

x = 0 or 2x â€“ 5 = 0

x = 0 or 2x = 5

x = 0 or x = 5/2

âˆ´ Value of x = 0, 5/2

**(ii) **xÂ² â€“ 2x = 48

Let us simplify the given equation,

By factorizing, we get

x^{2} â€“ 2x â€“ 48 = 0

x^{2} â€“ 8x+ 6x â€“ 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

âˆ´ Value of x = 8, -6

**3.** **(i) 6 + x = xÂ²**

**(ii) 2xÂ² + 3x + 1= 0**

**Solution:**

**(i) **6 + x = xÂ²

Let us simplify the given equation,

6 + x â€“ x^{2} = 0

x^{2} â€“ x â€“ 6 = 0

By factorizing, we get

x^{2} â€“ 3x + 2x â€“ 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

âˆ´ Value of x = 3, -2

**(ii) **2xÂ² + 3x + 1= 0

Let us simplify the given equation,

By factorizing, we get

2x^{2} â€“ 2x â€“ x + 1 = 0

2x(x – 1) â€“ 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 = Â½

âˆ´ Value of x = 1, Â½

**4. (i) 3xÂ² = 2x + 8(ii) 4xÂ² + 15 = 16x **

**Solution:**

**(i) **3xÂ² = 2x + 8

Let us simplify the given equation,

3x^{2} â€“ 2x â€“ 8 = 0

By factorizing, we get

3x^{2} â€“ 6x + 4x â€“ 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

âˆ´ Value of x = 2 or -4/3

**(ii) **4xÂ² + 15 = 16x

Let us simplify the given equation,

4x^{2} â€“ 16x + 15 = 0

By factorizing, we get

4x^{2} – 6x â€“ 10x + 15 = 0

2x(2x – 3) â€“ 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

âˆ´ Value of x = 3/2 or 5/2

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

**(ii) (x + 3) (x â€“ 3) = 40**

**Solution:**

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

Let us simplify the given equation,

2x^{2} + 5x â€“ 25 = 0

By factorizing, we get

2x^{2} + 10x â€“ 5x â€“ 25 = 0

2x(x + 5) â€“ 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

âˆ´ Value of x = -5, 5/2

**(ii) **(x + 3) (x â€“ 3) = 40

Let us simplify the given equation,

x^{2} â€“ 3x + 3x â€“ 9 = 40

x^{2} â€“ 9 â€“ 40 = 0

x^{2} â€“ 49 = 0

x^{2} = 49

x = âˆš49

= **Â±** 7

âˆ´ Value of x = 7, -7

**6. (i) (2x + 3) (x â€“ 4) = 6(ii) (3x + 1) (2x + 3) = 3**

**Solution:**

**(i) **(2x + 3) (x â€“ 4) = 6

Let us simplify the given equation,

2x^{2} â€“ 8x + 3x â€“ 12 â€“ 6 = 0

2x^{2} â€“ 5x â€“ 18 = 0

By factorizing, we get

2x^{2} â€“ 9x + 4x â€“ 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

âˆ´ Value of x = 9/2, -2

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

Let us simplify the given equation,

6x^{2} + 9x + 2x + 3 â€“ 3 = 0

6x^{2} + 11x = 0

x(6x + 11) = 0

So,

x = 0 or 6x + 11 = 0

x = 0 or 6x = -11

x = 0 or x = -11/6

âˆ´ Value of x = 0, -11/6

**7. (i) 4xÂ² + 4x + 1 = 0(ii) (x â€“ 4)Â² + 5Â² = 132**

**Solution:**

**(i) **4xÂ² + 4x + 1 = 0

Let us simplify the given equation,

By factorizing, we get

4x^{2} + 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

âˆ´ Value of x = -1/2, -1/2

**(ii) **(x â€“ 4)Â² + 5Â² = 13^{2}

Let us simplify the given equation,

x^{2} + 16 â€“ 2(x) (4) + 25 â€“ 169 = 0

x^{2} â€“ 8x -128 = 0

By factorizing, we get

x^{2} â€“ 16x + 8x â€“ 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

âˆ´ Value of x = 16, -8

**8. (i) 21x ^{2} = 4 (2x + 1)**

**(ii) 2/3x ^{2} â€“ 1/3x â€“ 1 = 0**

**Solution:**

**(i) **21x^{2} = 4 (2x + 1)

Let us simplify the given equation,

21x^{2} = 8x + 4

21x^{2} â€“ 8x â€“ 4 = 0

By factorizing, we get

21x^{2} â€“ 14x + 6x â€“ 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

âˆ´ Value of x = 2/3 or -2/7

**(ii) **2/3x^{2} â€“ 1/3x â€“ 1 = 0

Let us simplify the given equation,

By taking 3 as LCM and cross multiplying

2x^{2} â€“ x â€“ 3 = 0

By factorizing, we get

2x^{2} â€“ 3x + 2x â€“ 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

âˆ´ Value of x = 3/2, -1

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

**(ii) x + 1/x = 2 Â½ **

**Solution:**

**(i) **6x + 29 = 5/x

Let us simplify the given equation,

By cross multiplying, we get

6x^{2} + 29x â€“ 5 = 0

By factorizing, we get

6x^{2} + 30x â€“ x â€“ 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

âˆ´ Value of x = -5, 1/6

**(ii) **x + 1/x = 2 Â½** **

x + 1/x = 5/2

Let us simplify the given equation,

By taking LCM

x^{2} + 1 = 5x/2

By cross multiplying,

2x^{2} + 2 â€“ 5x = 0

2x^{2} â€“ 5x + 2 = 0

By factorizing, we get

2x^{2} â€“ x â€“ 4x + 2 = 0

x(2x – 1) â€“ 2 (2x â€“ 1) = 0

(2x – 1) (x – 2) = 0

So,

(2x – 1) = 0 or (x – 2) = 0

2x = 1 or x = 2

x = Â½ or x = 2

âˆ´ Value of x = Â½, 2

**10. (i) 3x â€“ 8/x = 2**

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

**Solution:**

**(i) **3x â€“ 8/x = 2

Let us simplify the given equation,

By taking LCM and cross multiplying,

3x^{2} â€“ 8 = 2x

3x^{2} â€“ 2x â€“ 8 = 0

By factorizing, we get

3x^{2} â€“ 6x + 4x â€“ 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

âˆ´ 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

x^{2} + 27 = 12x

x^{2} â€“ 12x + 27 = 0

By factorizing, we get

x^{2} â€“ 3x â€“ 9x + 27 = 0

x (x – 3) â€“ 9 (x – 3) = 0

(x – 3) (x – 9) = 0

So,

(x – 3) = 0 or (x – 9) = 0

x = 3 or x = 9

âˆ´ Value of x = 3, 9

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

**(ii) 1/(x + 2) + 1/x = Â¾**

**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)

3x^{2} â€“ 7x â€“ 3x + 7 = 2x^{2} + 2x â€“ 5x â€“ 5

3x^{2} â€“ 10x + 7 â€“ 2x^{2} +3x + 5 = 0

x^{2} â€“ 7x + 12 = 0

By factorizing, we get

x^{2} â€“ 4x â€“ 3x + 12 = 0

x (x – 4) â€“ 3 (x – 4) = 0

(x – 4) (x – 3) = 0

So,

(x – 4) = 0 or (x – 3) = 0

x = 4 or x = 3

âˆ´ Value of x = 4, 3

**(ii) **1/(x + 2) + 1/x = Â¾

Let us simplify the given equation,

By taking x(x + 2) as LCM

(x+x+2)/x(x + 2) = Â¾

By cross multiplying,

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

8x + 8= 3x^{2} + 6x

3x^{2} + 6x â€“ 8x â€“ 8 = 0

3x^{2} â€“ 2x â€“ 8 = 0

By factorizing, we get

3x^{2} â€“ 6x + 4x â€“ 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

âˆ´ Value of x = 2, -4/3

**12. (i) 8/(x + 3) â€“ 3/(2 – x) = 2**

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

**Solution:**

**(i) **8/(x + 3) â€“ 3/(2 – x) = 2

Let us simplify the given equation,

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

[8(2-x) â€“ 3(x+3)] / (x+3) (2-x) = 2 [16 â€“ 8x â€“ 3x â€“ 9] / [2x â€“ x^{2}+ 6 â€“ 3x] = 2 [-11x + 7] = 2(-x

^{2}â€“ x + 6)

7 â€“ 11x = -2x^{2} â€“ 2x + 12

2x^{2} + 2x â€“ 11 x â€“ 12 + 7 = 0

2x^{2} â€“ 9x â€“ 5 = 0

By factorizing, we get

2x^{2} â€“ 10x + x â€“ 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

âˆ´ 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) = 13/66[x^{2} + x^{2} + x + x + 1] = 13x(x + 1)

6[2x^{2} + 2x + 1] = 13x^{2} + 13x

12x^{2} + 12x + 6 â€“ 13x^{2} â€“ 13x = 0

-x^{2} â€“ x + 6 = 0

x^{2} + x â€“ 6 = 0

By factorizing, we get

x^{2} + 3x â€“ 2x â€“ 6 = 0

x (x + 3) â€“ 2 (x + 3) = 0

(x + 3) (x – 2) = 0

So,

(x + 3) = 0 or (x – 2) = 0

x = -3 or x = 2

âˆ´ Value of x = -3, 2

Chapter Test

**Solve the following (1 to 3) equations:**

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

**(ii) 3x ^{2} â€“ 4x â€“ 4 = 0.**

**Solution:**

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

We can write it as

2x^{2} + 5x â€“ 3 = 0

By further calculation

2x^{2} + 6x â€“ x â€“ 3 = 0

By taking out the common terms

2x (x + 3) â€“ 1 (x + 3) = 0

So we get

(x + 3) (2x â€“ 1) = 0

Here

x + 3 = 0 then x = – 3

2x â€“ 1 = 0 then 2x = 1 where x = Â½

Therefore, x = – 3, Â½.

(ii) 3x^{2} â€“ 4x â€“ 4 = 0

We can write it as

3x^{2} â€“ 6x + 2x â€“ 4 = 0

By taking out the common terms

3x (x â€“ 2) + 2 (x â€“ 2) = 0

So we get

(x â€“ 2) (3x + 2) = 0

Here

x â€“ 2 = 0 then x = 2

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

Therefore, x = 2, – 2/3.

**2. (i) 4x ^{2} â€“ 2x + Â¼ = 0**

**(ii) 2x ^{2} + 7x + 6 = 0.**

**Solution:**

(i) 4x^{2} â€“ 2x + Â¼ = 0

Multiply the equation by 4

16x^{2} â€“ 8x + 1 = 0

We can write it as

16x^{2} â€“ 4x â€“ 4x + 1 = 0

Taking out the common terms

4x (4x â€“ 1) â€“ 1 (4x â€“ 1) = 0

So we get

(4x â€“ 1) (4x â€“ 1) = 0

(4x â€“ 1)^{2} = 0

Here

4x â€“ 1 = 0

4x = 1

By division

x = Â¼, Â¼

(ii) 2x^{2} + 7x + 6 = 0

We can write it as

2x^{2} + 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 â€“ 1)/ (x â€“ 2) + (x â€“ 3)/ (x â€“ 4) = 3 1/3**

**(ii) 6/x â€“ 2/(x â€“ 1) = 1/(x â€“ 2).**

**Solution:**

(i) (x â€“ 1)/ (x â€“ 2) + (x â€“ 3)/ (x â€“ 4) = 3 1/3

By taking LCM

[(x â€“ 1) (x â€“ 4) + (x â€“ 2) (x â€“ 3)]/ (x â€“ 2) (x â€“ 4) = 10/3By further calculation

(x^{2} â€“ 5x + 4 + x^{2} â€“ 5x + 6)/ (x^{2} â€“ 6x + 8) = 10/3

So we get

(2x^{2} â€“ 10x + 10)/ (x^{2} â€“ 6x + 8) = 10/3

By cross multiplication

10x^{2} â€“ 60x + 80 = 6x^{2} â€“ 30x + 30

By further simplification

10x^{2} â€“ 60x + 80 â€“ 6x^{2} + 30x â€“ 30 = 0

So we get

4x^{2}Â â€“ 30x + 50 = 0

Dividing by 2

2x^{2} â€“ 15x + 25 = 0

It can be written as

2x^{2} â€“ 10x â€“ 5x + 25 = 0

Taking out the common terms

2x (x â€“ 5) â€“ 5 (x â€“ 5) = 0

(x â€“ 5) (2x â€“ 5) = 0

Here

x â€“ 5 = 0 then x = 5

2x â€“ 5 = 0 then 2x = 5 where x = 5/2

Therefore, x = 5, 5/2.

(ii) 6/x â€“ 2/(x â€“ 1) = 1/(x â€“ 2)

Taking LCM

(6x â€“ 6 â€“ 2x)/ x (x â€“ 1) = 1/ (x â€“ 2)

By further calculation

(4x â€“ 6)/ (x^{2} â€“ x) = 1/(x â€“ 2)

By cross multiplication

4x^{2} â€“ 8x â€“ 6x + 12 = x^{2} â€“ x

So we get

4x^{2} â€“ 14x + 12 â€“ x^{2} + x = 0

3x^{2} â€“ 13x + 12 = 0

3x^{2} â€“ 4x â€“ 9x + 12 = 0

Taking out the common terms

x (3x â€“ 4) â€“ 3 (3x â€“ 4) = 0

(3x â€“ 4) (x â€“ 3) = 0

Here

3x â€“ 4 = 0 then 3x = 4 where x = 4/3

x â€“ 3 = 0 then x = 3

Therefore, x = 3, 4/3.