*ICSE Maths Quadratic Equations is the seventh chapter of the ML Aggarwal book for class 9 . Here are the chapter 7 ML Aggarwal Class 9 Solution of Exercise.*

**Contents**Show

## ICSE Class 9 Maths Quadratic Equations

In ML Aggarwal Maths book for class 9 Chapter 7 is Quadratic Equations. In ICSE Quadratic Equation chapter is most important. This chapter consists of the following topics Algebra – Quadratic Equation, Factorisation method, etc The solution of Chapter 7 Quadratic Equations Maths ML Aggarwal Class are provided here.

ML Aggarwal Class 9 Solution Maths Exercise 7.1 ( Quadratic Equation)

## ML Aggarwal Class 9 Maths Chapter 7 : Quadratic Equation ( Exercise Solutions)

Chapter | Chapter 7 : Quadratic Equations |

Exercise | Exercise 7.1 ( Quadratic Equations) |

Class | 9 ( Class IX ) |

Total Questions | 15 Questions Practice Question : 12 Question Chapter Test Questions : 3 Questions |

Text Book | ML Aggarwal for Class 9 Maths [ Understanding ICSE Mathematics] |

### ICSE Maths Quadratic Equation Class 9 Solution ( ML Aggarwal)

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

**ML Aggarwal Class 9 Solutions Chapter 7 Question 1 :(i) x ^{2} – 11x + 30 = 0 (ii) 4x^{2} – 25 = 0Answer : **

**(i) x**

^{2}-11x + 30 = 0Doing Factorisation,

we get,

x

^{2}-5x -6x + 30 = 0

x(x-5) -6 ( x-5) = 0

Taking (X-5) as common,

(x-5) (x-6) = 0 …..(1)

Now from equation (1) ,

we get

either (x-5) =0

Or (x-6) = 0

So , if (x-5 ) = 0 , then x = 5.

If (x-6 ) = 0 , then x= 6.

Values of x are : 5 , 6

**(ii) 4x**

^{2}-25 = 0Simplifying the given equation ,

we get,

4x

^{2}= 25

x

^{2}= 25/4

Taking sqr root on both side,

we get

x = \( \pm\sqrt{\frac{25}{4}}\ =\ \pm\frac{5}{2} \)

Values of x = + 5/2 , -5/2

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 2 :****(i) 2x ^{2} – 5x = 0 (ii) x^{2} – 2x = 48 **

**Answer :**

(i) 2x

^{2}-5x = 0

Taking ‘x’ as common ,

we get ,

x ( 2x – 5) = 0 ……1

From equation 1 we came to known that,

Either x =0

Or 2x – 5 = 0

If 2x-5=0 , then

2x = 5

x = 5/2

So values of x are 0 , 5/2

**(ii) x**

^{2}-2x = 48The given equation can be rewritten as

x

^{2}– 2x -48 = 0

Doing factorisation,

we get,

x

^{2}– 8x +6x -48 = 0

x(x-8) +6(x-8) = 0

(x+6) ( x-8) = 0

Either ( x+6) = 0

Or ( x-8) = 0

If x+6 = 0,

x = -6

If x-8 = 0, then

x=8

Values of x = -6 , 8

**ML Aggarwal Class 9 Solutions Chapter 7 Question 3 : **(i) 6 + x = x

^{2}

(ii) 2x

^{2}+3x +1 =0

Answer :

**(i) 6 +x = x**

^{2}Rearranging equation as follows ,

x

^{2}-x – 6 = 0

Doing factorisation, we get

x

^{2}– 3x +2x – 6 = 0

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

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

Either (x-3) = 0

or (x+ 2) = 0

If (x-3) = 0,then x =3

If x+2 = 0 , then x =-2

Value of x = 3, -2

**(ii) 2x**

^{2}+3x+1=0Doing factorisation of the given equation , we get

2x

^{2}+ 3x +1 = 0

2x

^{2}+2x +x + 1 = 0

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

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

Therefore,

If ( x+1) = 0 then, x = -1

If (2x+1) = 0 then x= -1/2

Value of x = -1, -1/2

**ML Aggarwal Class 9 Solutions Chapter 7 Question 4 :(i) 3x ^{2} = 2x + 8 (ii) 4x^{2} + 15 = 16xAnswer : (i) 3x^{2} = 2x + 8 **

The above equation can be rewritten as:

3x

^{2}-2x – 8 =0

Doing Factorisation ,

3x

^{2}– 6x + 4x – 8 = 0

3x (x-2) + 4 (x-2) = 0

(3x +4) ( x-2) = 0

If (x-2) = 0 then , x =2

If (3x + 4 ) =0 then, x = -4/3

Value of x =2 , -4/3

**(ii) 4x**

^{2}+15 = 16xThe given equation can be rewritten as

4x

^{2}-16x +15 = 0

Doing factorisation,

4x

^{2}-6x -10x +15 = 0

2x(2x-3) -5 ( 2x-3) = 0

(2x-3) (2x-5) = 0

If

**(2x-3) = 0 then, x = 3/2**

If (2x-5) = 0 then , x = 5/2

Therefore the value of

x = 3/2 and 5/2

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 5 :**

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

(ii) ( x + 3) ( x – 3) = 40**Answer : ****(i) x (2x+5) = 25 **

2x^{2}+5x = 25

2x^{2} +5x -25 =0

2x^{2} +10x -5x -25 = 0

2x(x+5) -5(x+5) =0

(x + 5) (2x – 5) = 0

If ( x + 5) = 0 then x = -5

If (2x – 5) = 0 then , x = 5/2

The values of x are -5, 5/2 **(ii)( x + 3) ( x – 3) = 40**

x^{2} – 3x + 3x – 9 = 40

x^{2} – 9 – 40 = 0

x^{2} – 49 = 0

X^{2} = 49** **

X = ±7

The values of x are -7 and -7.

**ML Aggarwal Class 9 Solutions Chapter 7 Question 6 :(i) ( 2x + 3) ( x – 4) = 6 (ii) ( 3x + 1) ( 2x + 3) = 3 Answer : (i) ( 2x + 3) ( x – 4) = 6 **

The equation can be simplified as under :

2x

^{2}– 8x + 3x – 12 – 6 = 0

2x

^{2}– 5x -18 = 0

Now doing factorising,

we get

2x

^{2}– 9x + 4x – 18 = 0

x(2x – 9) + 2(2x-9) =0

(2x – 9) (x +2) = 0

Now,

If (2x – 9 ) = 0 then , x = 9/2

If (x + 2) = 0 then, x= -2

The value of ‘x’ are 9/2 and -2.

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

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

The equation can be simplified as follows

6x

^{2}+ 9x + 2x + 3 -3 = 0

6x

^{2}+ 11x = 0

x(6x + 11) = 0

Therefore ,

x = 0

Or (6x + 11) = 0 , then , x = -11/6

The value of x are 0 and -11/6

**ML Aggarwal Class 9 Solutions Chapter 7 Question 7 :(i) 4x ^{2} + 4x + 1 = 0 (ii) ( x – 4)^{2} + 5^{2} = 13^{2}Answer : (i) 4x^{2} + 4x + 1 = 0 **

Doing factorising,

we get

4x

^{2}+ 2x + 2x + 1 = 0

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

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

Therefore,

(2x+1) = 0 then , x = -1/2

The values of x are -1/2, -1/2.

(ii) ( x – 4)

(ii) ( x – 4)

^{2}+ 5^{2}= 13^{2}x

^{2}+ 16 -2(x) *( 4) + 25 – 169 = 0 ( using algebraic identity, (a+b)

^{2}= a

^{2}+ b

^{2}+ 2ab )

x

^{2}– 8x – 128 = 0

Doing Factorisation,

we get

x

^{2}– 16x + 8x – 128 = 0

x(x – 16) + 8(x – 16) = 0

( x – 16) ( x + 8) = 0

Therefore,

If (x-16)=0 then, x = 16

If (x+8) = 0 then , x = -8

The values of ‘x’ are 16 and -8

**ML Aggarwal Class 9 Solutions Chapter 7 Question 8 :(i) 21x ^{2} = 4( 2x + 1) (ii) \( \frac{2}{3x^{2\ }}-\frac{\ \ 1}{3x}-1=0 \)Answer : (i) 21x^{2} = 4( 2x + 1) **

The equation can be simplified as follows:

21x

^{2}= 8x +4

21x

^{2}– 8x – 4 = 0

Using factorisation method ,

we get,

21x

^{2}– 14x + 6x – 4 = 0

7x(3x – 2) +2( 3x – 2) = 0

(3x – 2) ( 7x + 2) = 0

If (3x-2) =0 then, x = 2/3

If ( 7x+2) = 0 , then x = -2/7

The values of ‘x’ are -2/7 and 2/3.

(ii) \( \frac{2}{3x^{2\ }}-\frac{\ \ 1}{3x}-1=0 \)

(ii) \( \frac{2}{3x^{2\ }}-\frac{\ \ 1}{3x}-1=0 \)

The equation can be simplified as follows :

The L.C.M. of 3x

^{2}and 3x and 1 = 3x

^{2}.

Solving further,

we get ,

\( \frac{2-x-3x^2}{3x^2}\ \ =\ 0\ \)

Now Equation can be simplified further,

2 – x – 3x

^{2}= 0

-3x

^{2}– x +2 =0

Multiplying by minus ( -) on both side ,

3x

^{2}+ x -2 = 0

Using Factorization method,

3x

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

3x( x +1 ) -2 ( x +1) = 0

(3x-2) ( x+1) = 0

If ( 3x-2) = 0 , then x =2/3

If (x+1) =0 , then x = -1

The value of ‘x’ are 2/3 and -1

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 9 :****(i) 6x + 29 = 5/x (ii) x + 1/𝑥 = 2 \( \ frac{1}{2} \)Answer : **

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

The equation can be simplified as follows :

6x

^{2}+ 29x – 5 = 0

Now , using the factorisation method ,

we get,

6x

^{2}+ 30x – x – 5 = 0

6x (x + 5) -1 (x + 5) = 0

(x + 5) (6x – 1) = 0

If (x + 5) = 0 then, x =-5

If (6x – 1) = 0 , then x = 1/6

Therefore,

The value of ‘x’ are -5 and 1/6.

**(ii) x + 1/x = 2 ½**

The equation cab be simplified as follows :

x + 1/x = 5/2

x

^{2}+ 1 = 5x/2

2x

^{2}+ 2 – 5x = 0

2x

^{2}– 5x + 2 = 0

Now Using factorization method ,

we get,

2𝑥

^{2}– 𝑥 – 4𝑥 + 2 = 0

𝑥(2𝑥 – 1) – 2(2𝑥 – 1) = 0

(2𝑥 – 1) (𝑥 – 2) = 0

If (2𝑥 – 1) = 0 then , 𝑥 = 1/2

If (𝑥 – 2) = 0 then , 𝑥 = 2

The value of ‘𝑥’ are 1/2 and 2 .

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 10 :****(i) 3𝑥 – 8 /𝑥 = 2 (ii) 𝑥/3 + 9/ 𝑥 = 4**

**Answer :**

(i)

**)**3x – 8/x = 2

The above equation cab be simplified as follows :

3x

^{2}– 8 = 2x

3x

^{2}– 2x – 8 = 0

Using the factorisation method,

we get,

3x

^{2}– 6x + 4x – 8 = 0

3x(x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

If, (x – 2) = 0 , then x = 2

If (3x + 4) = 0 , then x = -4/3

There fore,

The value of ‘x’ are 2 and -4/3.

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

Simplifying the above equation using Cross multiplying

We get,

x

^{2}+ 27 = 12x

x

^{2}– 12x + 27 = 0

Using factorisation method ,

we get,

x

^{2}– 3x – 9x + 27 = 0

x (x – 3) – 9 (x – 3) = 0

(x – 3) (x – 9) = 0

If, (x – 3) = 0 , then x = 3

if (x – 9) = 0, then x = 9

Therefore,

The value of ‘x’ are 3 and 9.

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 11 :****(i) (𝑥−1) /(𝑥+1 )= (2𝑥−5)/ ( 3𝑥−7) (ii) 1/ (𝑥+2) + 1 /𝑥 = 3/4**

**Answer :**

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

Simplifying the given equation using cross multiplication method,

we get,

(x – 1) (3x – 7) = (2x – 5) (x + 1)

3x

^{2}– 7x – 3x + 7 = 2x

^{2}+ 2x – 5x – 5

3x

^{2}– 10x + 7 – 2x

^{2}+3x + 5 = 0

x

^{2}– 7x + 12 = 0

Using the factorization method ,

we get,

x

^{2}– 4x – 3x + 12 = 0

x (x – 4) – 3 (x – 4) = 0

(x – 4) (x – 3) = 0

If (x – 4) = 0 then , x = 4

if (x – 3) = 0 then, x = 3

Therefore,

The values of ‘x’ are 4 and 3 .

**(ii) 1/(x + 2) + 1/x = ¾**

Simplifying the given equation,

We get,

(x+x+2)/[x(x + 2)] = 3/4

4(2x + 2) = 3x(x + 2) …. Using the cross multiplication method

8x + 8= 3x

^{2}+ 6x

3x

^{2}+ 6x – 8x – 8 = 0

3x

^{2}– 2x – 8 = 0

3x

^{2}– 6x + 4x – 8 = 0 …. Using factorisation Method

3x(x – 2) + 4 (x – 2) = 0

(x – 2) (3x + 4) = 0

If (x – 2) = 0, then x =2

If (3x + 4) =0 then, x =-4/3

Therefore,

The value of ‘x’ are 2 and -4/3

**ML Aggarwal Class 9 Solutions Chapter 7 ****Question 12 :****(i) 8 /(𝑥+3) – 3/( 2−𝑥) = 2 (ii) 𝑥 /(𝑥+1) + (𝑥+1)/ 𝑥 = \( 2\frac{1}{6} \) **

**Answer :**

**(i) 8 /(𝑥+3) – 3/( 2−𝑥) = 2**

The equation can be simplified as follows :

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

[8(2-x) – 3(x+3)] / (x+3) (2-x) = 2 …( LCM (x+3) ( 2-x) on LHS )

[16 – 8x – 3x – 9] / [2x – x

^{2}+ 6 – 3x] = 2

[-11x + 7] = 2(-x

^{2}– x + 6)

7 – 11x = -2x

^{2}– 2x + 12

2x

^{2}+ 2x – 11 x – 12 + 7 = 0

2x

^{2}– 9x – 5 = 0

2x

^{2}– 10x + x – 5 = 0 … (Using factorisation method)

2x (x – 5) + 1 (x – 5) = 0

(x – 5) (2x + 1) = 0

If (x – 5) = 0 then, x = 5

If (2x + 1) = 0 , then x = -1/2

Therefore,

The value of ‘x’ are 5 and -1/2

**(ii) 𝑥 /(𝑥+1) + (𝑥+1)/ 𝑥 = \( 2\frac{1}{6} \)**

The above equation cab be simplified as follows:

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

By taking x(x+1) as LCM

[x(x) + (x+1) (x+1)] / x(x + 1) = 13/6 … ( LCM x and (x+1) = x(x+1) on LHS )

6[x

^{2}+ x

^{2}+ x + x + 1] = 13x(x + 1)

6[2x

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

^{2}+ 13x

12x

^{2}+ 12x + 6 – 13x

^{2}– 13x = 0

-x

^{2}– x + 6 = 0

x

^{2}+ x – 6 = 0 … ( Multiplying by ‘ – ‘ on both sides )

x

^{2}+ 3x – 2x – 6 = 0 …( Using the factorisation method )

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

If (x + 3) = 0 , then x = 3

If (x – 2) = 0, then x = 2

Therefore,

The value of ‘x’ are -3 and 2.

**C hapter Test**

**Question 1 :****(i) x (2x + 5) = 3 (ii) 3x ^{2} – 4x – 4 = 0**

**Answer :**

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

Simplifying the equation,

we get,

2x

^{2}+ 5x – 3 = 0

2x

^{2}+ 6x – x – 3 = 0 …( Using Factorisation metho)

2x (x + 3) – 1 (x + 3) = 0

(x + 3) (2x – 1) = 0

If (x + 3 = 0 then x = – 3

If (2x – 1) = 0 then x = 1/2

Therefore,

The value of ‘x’ are -3 and 1/2

**(ii) 3x**

^{2}– 4x – 4 = 0Using the factorisation method,

we get

3x

^{2}– 6x + 2x – 4 = 0

3x (x – 2) + 2 (x – 2) = 0

(x – 2) (3x + 2) = 0

If (x – 2) = 0 then, x = 2

If ( 3x +2) = 0 then, x = -2/3

Therefore,

The values of ‘x’ are 2 and -2/3

**Question 2:****(i) 4x ^{2} – 2x + 1/4 = 0 (ii) 2x^{2} + 7x + 6 = 0Answer : **

**(i) 4x**

^{2}– 2x + 1/4 = 0Taking LCM as 4 and doing further calculations ,

(16x

^{2}– 8x + 1) / 4 = 0

16x

^{2}– 8x + 1= 0

16x

^{2}– 4x – 4x + 1 = 0 .. (Using factorisation method )

4x (4x – 1) – 1 (4x – 1) = 0

(4x – 1) (4x – 1) = 0

If ( 4x-1) = 0 , then x =1/4

Therefore

The values of ‘x’ are 1/4 and 1/4 .

**(ii) 2x**

^{2}+ 7x + 6 = 02x

^{2}+ 4x + 3x + 6 = 0

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

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

If ( x + 2) = 0 then x = – 2

If (2x + 3) = 0 then x = – 3/2

Therefore,

The values of ‘x’ are -2 and -3/2

**Question 3:** **(i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = \( 3\frac{1}{3}** \) **(ii) 6/x – 2/(x – 1) = 1/(x – 2).** **Answer :**

(i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = \( 3\frac{1}{3} \)

[(x – 1) (x – 4) + (x – 2) (x – 3)]/ (x – 2) (x – 4) = 10/3 ..( LCM of (x-2) and (x-4) = (x-2) (x-4)

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

(2x^{2} – 10x + 10)/ (x^{2} – 6x + 8) = 10/3

10x^{2} – 60x + 80 = 6x^{2} – 30x + 30

10x^{2} – 60x + 80 – 6x^{2} + 30x – 30 = 0

4x^{2} – 30x + 50 = 0

2x^{2} – 15x + 25 = 0 … (Multiplying 1/2 on both side )

2x^{2} – 10x – 5x + 25 = 0 … (Using the factorisation method)

2x (x – 5) – 5 (x – 5) = 0

(x – 5) (2x – 5) = 0

If (x – 5) = 0 then, x = 5

If (2x – 5) = 0 then x = 5/2 **(ii) 6/x – 2/(x – 1) = 1/(x – 2)**

(6x – 6 – 2x)/ x (x – 1) = 1/ (x – 2) … ( On LHS LCM of x and (x-1) is x(x-1) )

(4x – 6)/ (x^{2} – x) = 1/(x – 2)

4x^{2} – 8x – 6x + 12 = x^{2} – x

4x^{2} – 14x + 12 – x^{2} + x = 0

x^{2} – 13x + 12 = 0

3x^{2} – 4x – 9x + 12 = 0 … (Using the factorisation method)

x (3x – 4) – 3 (3x – 4) = 0

(3x – 4) (x – 3) = 0

If (3x-4) = 0 then, x =4/3

If (x – 3) = 0 then, x = 3

Therefore,

x = 3, 4/3.