1. Solve the following quadratic equations byfactorization method:(i)

1.	1. Solve the following quadratic equations byfactorization method:(i) 3x2 2x +8(iii) x2 -5x-36 0 (iv) (2x+3)2 -16(ii) 2x2 + 5x + 2 = 0.
Answer» i) 3x² = 2x+8 => 3x²-2x-8 = 0 => 3x²-6x+4x-8 = 0 => 3x(x-2)+4(x-2) = 0 => (3x+4)(x-2) = 0 , the value of x = x = -4/3 , 2 ii) 2x²+5x+2 = 0 => 2x²+4x+x+2 =0 => 2x(x+2) +1(x+2) = 0 => (2x+1)(x+2) =0 so the value of x = -1/2 , -2 iii) x²-5x-36 = 0 => x²-9x+4x-36 =0 => x(x-9)+4(x-9) =. => (x+4)(x-9) = 0 so x = 9,-4. iv) (2x+3)² = 16 => 2x+3 = √16 = +4,-4 => 2x = +4-3 , -4-3 => x = 1/2 , -7/2.

1. Solve the following quadratic equations byfactorization method:(i) 3x2 2x +8(iii) x2 -5x-36 0 (iv) (2x+3)2 -16(ii) 2x2 + 5x + 2 = 0.

Answer»

i) 3x² = 2x+8 => 3x²-2x-8 = 0 => 3x²-6x+4x-8 = 0 => 3x(x-2)+4(x-2) = 0 => (3x+4)(x-2) = 0 , the value of x = x = -4/3 , 2

ii) 2x²+5x+2 = 0 => 2x²+4x+x+2 =0 => 2x(x+2) +1(x+2) = 0 => (2x+1)(x+2) =0 so the value of x = -1/2 , -2

iii) x²-5x-36 = 0 => x²-9x+4x-36 =0 => x(x-9)+4(x-9) =. => (x+4)(x-9) = 0 so x = 9,-4.

iv) (2x+3)² = 16 => 2x+3 = √16 = +4,-4 => 2x = +4-3 , -4-3 => x = 1/2 , -7/2.