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Mathematics

Use the substitution y = 3x + 1 to solve for x :

5(3x + 1)2 + 6(3x + 1) - 8 = 0.

Quadratic Equations

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Answer

Given,

5(3x+1)2+6(3x+1)−8=0⇒5y2+6y−8=0 (Putting 3x + 1 = y) ⇒5y2+10y−4y−8=0⇒5y(y+2)−4(y+2)=0⇒(5y−4)(y+2)=0 (Factorising left side) ⇒5y−4=0 or y+2=0 (Zero-product rule) ⇒y=45 or y=−2When y=45,3x+1=45⇒3x=45−1⇒3x=−15⇒x=−115When y=−2,3x+1=−2⇒3x=−3⇒x=−15(3x + 1)^2 + 6(3x + 1) - 8 = 0 \\[1em] \Rightarrow 5y^2 + 6y - 8 = 0 \text{ (Putting 3x + 1 = y) } \\[1em] \Rightarrow 5y^2 + 10y - 4y - 8 = 0 \\[1em] \Rightarrow 5y(y + 2) - 4(y + 2) = 0 \\[1em] \Rightarrow (5y - 4)(y + 2) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow 5y - 4 = 0 \text{ or } y + 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow y = \dfrac{4}{5} \text{ or } y = -2 \\[1em] \text{When } y = \dfrac{4}{5}, 3x + 1 = \dfrac{4}{5} \\[1em] \Rightarrow 3x = \dfrac{4}{5} - 1 \\[1em] \Rightarrow 3x = -\dfrac{1}{5} \\[1em] \Rightarrow x = -\dfrac{1}{15} \\[1em] \text{When } y = -2 , 3x + 1 = -2 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1

Hence, the roots of given equation are −115-\dfrac{1}{15} , -1.

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