ML Aggarwal Solutions for Class 9 Maths Chapter 7- Quadratic Equations

Class 9: Maths Chapter 7 solutions. Complete Class 9 Maths Chapter 7 Notes.

ML Aggarwal Solutions for Class 9 Maths Chapter 7- Quadratic Equations

ML Aggarwal 9th Maths Chapter 7, Class 9 Maths Chapter 7 solutions

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.

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ML Aggarwal Solutions for Class 9 Maths Chapter 7- Quadratic Equations

Download PDF: ML Aggarwal Solutions for Class 9 Maths Chapter 7- Quadratic Equations PDF

Chapterwise ML Aggarwal Solutions for Class 9 Maths :

• Chapter 1- Rational and Irrational Numbers
• Chapter 2- Compound Interest
• Chapter 3- Expansions
• Chapter 4- Factorization
• Chapter 5- Simultaneous Linear Equations
• Chapter 6- Problems on Simultaneous Linear Equations
• Chapter 7- Quadratic Equations
• Chapter 8- Indices
• Chapter 9- Logarithms
• Chapter 10- Triangles
• Chapter 11- Mid Point Theorem
• Chapter 12- Pythagoras Theorem
• Chapter 13- Rectilinear Figures
• Chapter 14- Theorems on Area
• Chapter 15- Circle
• Chapter 16- Mensuration
• Chapter 17- Trigonometric Ratios
• Chapter 18- Trigonometric Ratios and Standard Angles
• Chapter 19- Coordinate Geometry
• Chapter 20- Statistics

About ML Aggarwal

M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.