## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2

Solve the following equations (1 to 24) by factorization:

Question 1.

(i) 4x² = 3x

(ii) \(\frac { { x }^{ 2 }-5x }{ 2 } =0\)

Solution:

Question 2.

(i) (x – 3) (2x + 5) = 0

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

Solution:

Question 3.

(i) x² – 3x – 10 = 0

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

Solution:

Question 4.

(i) 3x² – 5x – 12 = 0

(ii) 21x² – 8x – 4 = 0

Solution:

Question 5.

(i) 3x² = x + 4

(ii) x(6x – 1) = 35

Solution:

Question 6.

(i) 6p² + 11p – 10 = 0

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

Solution:

Question 7.

(i) (x – 4)² + 5² = 13²

(ii) 3(x – 2)² = 147

Solution:

Question 8.

(i) \(\\ \frac { 1 }{ 7 } \)(3x – 5)² = 28

(ii) 3(y² – 6) = y(y + 7) – 3

Solution:

Question 9.

x² – 4x – 12 = 0,when x ∈ N

Solution:

Question 10.

2x² – 8x – 24 = 0 when x ∈ I

Solution:

Question 11.

5x² – 8x – 4 = 0 when x ∈ Q

Solution:

Question 12.

2x² – 9x + 10 = 0,when

(i) x ∈ N

(ii) x ∈ Q

Solution:

Question 13.

(i) a²x² + 2ax + 1 = 0, a≠0

(ii) x² – (p + q)x + pq = 0

Solution:

Question 14.

a²x² + (a² + b²)x + b² = 0, a≠0

Solution:

Question 15.

(i) √3x² + 10x + 7√3 = 0

(ii) 4√3x² + 5x – 2√3 = 0

Solution:

Question 16.

(i) x² – (1 + √2)x + √2 = 0

(ii) \(x+ \frac { 1 }{ x } \) = \(2 \frac { 1 }{ 20 } \)

Solution:

Question 17.

(i) \(\frac { 2 }{ { x }^{ 2 } } -\frac { 5 }{ x } +2=0,x\neq 0 \)

(ii)\(\frac { { x }^{ 2 } }{ 15 } -\frac { x }{ 3 } -10=0 \)

Solution:

Question 18.

(i) \(3x-\frac { 8 }{ x } =2 \)

(ii) \(\frac { x+2 }{ x+3 } =\frac { 2x-3 }{ 3x-7 } \)

Solution:

Question 19.

(i) \(\frac { 8 }{ x+3 } -\frac { 3 }{ 2-x } =2 \)

(ii) \(\frac { x }{ x-1 } +\frac { x-1 }{ x } =2\frac { 1 }{ 2 } \)

Solution:

Question 20.

(i) \(\frac { x }{ x+1 } +\frac { x+1 }{ x } =\frac { 34 }{ 15 } \)

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

Solution:

Question 21.

(i) \(\frac { 1 }{ x-3 } -\frac { 1 }{ x+5 } =\frac { 1 }{ 6 } \)

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

Solution:

Question 22.

(i) \(\frac { a }{ ax-1 } +\frac { b }{ bx-1 } =a+b,a+b\neq 0,ab\neq 0\)

(ii) \(\frac { 1 }{ 2a+b+2x } =\frac { 1 }{ 2a } +\frac { 1 }{ b } +\frac { 1 }{ 2x } \)

Solution:

Question 23.

\(\frac { 1 }{ x+6 } +\frac { 1 }{ x-10 } =\frac { 3 }{ x-4 } \)

Solution:

Question 24.

(i) \(\sqrt { 3x+4 } =x\)

(ii) \(\sqrt { x(x-7) } =3\sqrt { 2 } \)

Solution:

Question 25.

Use the substitution y = 3x + 1 to solve for x : 5(3x + 1 )² + 6(3x + 1) – 8 = 0

Solution:

Question 26.

Find the values of x if p + 1 =0 and x² + px – 6 = 0

Solution:

Question 27.

Find the values of x if p + 7 = 0, q – 12 = 0 and x² + px + q = 0,

Solution:

Question 28.

If x = p is a solution of the equation x(2x + 5) = 3, then find the value of p.

Solution: