Question

Solve 3(2x - 1)^2 + 4(2x - 1) - 4 = 0
find the solution set ?

Answer 1

$\huge\red{Solution❥}$

→3(2x-1)²+4(2x-1)-4=0

→ 3(4x²+1-4x)+8x-4-4=0

→ 12x²+3-12x+8x-8=0

→ 12x²-4x-5=0

→ 12x²-10x+6x-5=0

→ 2x(6x-5)+1(6x-5)=0

→ (2x+1)(6x-5)=0

→ x=-1/2 or 5/6

$\huge\pink{@ItzSiddhi}$

Answer 2

• 3(2x - 1)² + 4(2x - 1) - 4 = 0

We know that,

(a - b)² = a² + b² - 2ab

• 3((2x)² + 1² - 2 . 2x . 1) + 8x - 4 - 4 = 0

• 3(4x² + 1 - 4x) + 8x - 8 = 0

• 12x² + 3 - 12x + 8x - 8 = 0

Rearranging,

• 12x² - 12x + 8x - 8 + 3 = 0

• 12x² - 4x - 5 = 0

By PSF method,

P = - 5 * 12x² = - 60x²

S = - 4x

F = - 10x , + 6x

• 12x² + 6x - 10x - 5 = 0

• 6x(2x + 1) - 5(2x + 1) = 0

• (2x + 1)(6x - 5) = 0

• 2x + 1 = 0 , 6x - 5 = 0

• 2x = - 1 , 6x = 5

• x = $\sf \dfrac{- \: 1}{2}$ , x = $\sf \dfrac{5}{6}$

Therefore,

• $\bf{\bf{\sf x \: = \: \dfrac{- \: 1}{2} \: , \: \dfrac{5}{6}}}$