Question

Solve for x: 5(3x+1)² + 6(3x+1) -8 = 0​

Answer 1

Answer:

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Answer 2

given :-

5(3x+1)² + 6(3x+1) - 8 = 0​

⇒ 5[(3x)^2 + 2(3x)(1) + (1)^2] + 18x + 6 - 8 = 0

⇒ 5(9x^2 + 6x + 1) + 18x - 2 = 0

⇒ 45x^2 + 30x + 5 + 18x - 2 = 0

⇒ 45x^2 + 48x + 3 = 0

taking 3 as common,

⇒ 15x^2 + 16x + 1 = 0

⇒ 15x^2 + (15x + x) + 1 = 0

⇒ 15x^2 + 15x + x + 1 = 0

⇒ 15x(x + 1) + 1(x + 1) = 0

⇒ (x + 1) (15x + 1) = 0

• x = -1, x = -1/15

identity used :- (a + b)^2 = a^2 + 2ab + b^2