# ICSE Maths Quadratic Equation ML Aggarwal Class 9 Solution

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

## 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)

### 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) x2 – 11x + 30 = 0
(ii) 4x2 – 25 = 0

(i) x2 -11x + 30 = 0
Doing Factorisation,
we get,
x2 -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) 4x2 -25 = 0
Simplifying the given equation ,
we get,
4x2 = 25
x2 = 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) 2x2 – 5x = 0
(ii) x2 – 2x = 48

(i) 2x2 -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) x2 -2x = 48
The given equation can be rewritten as
x2 – 2x -48 = 0
Doing factorisation,
we get,
x2 – 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 = x2
(ii) 2x2 +3x +1 =0

(i) 6 +x = x2
Rearranging equation as follows ,
x2 -x – 6 = 0
Doing factorisation, we get
x2 – 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) 2x2 +3x+1=0
Doing factorisation of the given equation , we get
2x2 + 3x +1 = 0
2x2 +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) 3x2 = 2x + 8
(ii) 4x2 + 15 = 16x
(i) 3x2 = 2x + 8

The above equation can be rewritten as:
3x2 -2x – 8 =0
Doing Factorisation ,
3x2 – 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) 4x2 +15 = 16x
The given equation can be rewritten as
4x2 -16x +15 = 0
Doing factorisation,
4x2 -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
(i) x (2x+5) = 25
2x2+5x = 25
2x2 +5x -25 =0
2x2 +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
x2 – 3x + 3x – 9 = 40
x2 – 9 – 40 = 0
x2 – 49 = 0
X2 = 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
(i) ( 2x + 3) ( x – 4) = 6

The equation can be simplified as under :
2x2 – 8x + 3x – 12 – 6 = 0
2x2 – 5x -18 = 0
Now doing factorising,
we get
2x2 – 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

The equation can be simplified as follows
6x2 + 9x + 2x + 3 -3 = 0
6x2 + 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) 4x2 + 4x + 1 = 0
(ii) ( x – 4)2 + 52 = 132
(i) 4x2 + 4x + 1 = 0

Doing factorising,
we get
4x2 + 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)2 + 52 = 132

x2 + 16 -2(x) *( 4) + 25 – 169 = 0 ( using algebraic identity, (a+b)2 = a2 + b2 + 2ab )
x2 – 8x – 128 = 0
Doing Factorisation,
we get
x2 – 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) 21x2 = 4( 2x + 1)
(ii) $$\frac{2}{3x^{2\ }}-\frac{\ \ 1}{3x}-1=0$$
(i) 21x2 = 4( 2x + 1)

The equation can be simplified as follows:
21x2 = 8x +4
21x2 – 8x – 4 = 0
Using factorisation method ,
we get,
21x2 – 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) $$\frac{2}{3x^{2\ }}-\frac{\ \ 1}{3x}-1=0$$

The equation can be simplified as follows :
The L.C.M. of 3x2 and 3x and 1 = 3x2.
Solving further,
we get ,
$$\frac{2-x-3x^2}{3x^2}\ \ =\ 0\$$
Now Equation can be simplified further,
2 – x – 3x2 = 0
-3x2 – x +2 =0
Multiplying by minus ( -) on both side ,
3x2 + x -2 = 0
Using Factorization method,
3x2 + 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 $$\ frac{1}{2}$$

(i) 6x + 29 = 5/x
The equation can be simplified as follows :
6x2 + 29x – 5 = 0
Now , using the factorisation method ,
we get,
6x2 + 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
x2 + 1 = 5x/2
2x2 + 2 – 5x = 0
2x2 – 5x + 2 = 0
Now Using factorization method ,
we get,
2𝑥2 – 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

(i)3x – 8/x = 2

The above equation cab be simplified as follows :
3x2 – 8 = 2x
3x2 – 2x – 8 = 0
Using the factorisation method,
we get,
3x2 – 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,
x2 + 27 = 12x
x2 – 12x + 27 = 0
Using factorisation method ,
we get,
x2 – 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

(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)
3x2 – 7x – 3x + 7 = 2x2 + 2x – 5x – 5
3x2 – 10x + 7 – 2x2 +3x + 5 = 0
x2 – 7x + 12 = 0

Using the factorization method ,
we get,
x2 – 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= 3x2 + 6x
3x2 + 6x – 8x – 8 = 0
3x2 – 2x – 8 = 0
3x2 – 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 = $$2\frac{1}{6}$$

(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 – 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
2x2 – 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[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 … ( Multiplying by ‘ – ‘ on both sides )
x2 + 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.

Chapter Test

Question 1 :
(i) x (2x + 5) = 3
(ii) 3x2 – 4x – 4 = 0

(i) x (2x + 5) = 3
Simplifying the equation,
we get,
2x2 + 5x – 3 = 0
2x2 + 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) 3x2 – 4x – 4 = 0
Using the factorisation method,
we get
3x2 – 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) 4x2 – 2x + 1/4 = 0
(ii) 2x2 + 7x + 6 = 0

(i) 4x2 – 2x + 1/4 = 0
Taking LCM as 4 and doing further calculations ,
(16x2 – 8x + 1) / 4 = 0
16x2 – 8x + 1= 0
16x2 – 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) 2x2 + 7x + 6 = 0
2x2 + 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).

(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)
(x2 – 5x + 4 + x2 – 5x + 6)/ (x2 – 6x + 8) = 10/3
(2x2 – 10x + 10)/ (x2 – 6x + 8) = 10/3
10x2 – 60x + 80 = 6x2 – 30x + 30
10x2 – 60x + 80 – 6x2 + 30x – 30 = 0
4x2 – 30x + 50 = 0
2x2 – 15x + 25 = 0 … (Multiplying 1/2 on both side )
2x2 – 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)/ (x2 – x) = 1/(x – 2)
4x2 – 8x – 6x + 12 = x2 – x
4x2 – 14x + 12 – x2 + x = 0
x2 – 13x + 12 = 0
3x2 – 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.