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(B) Let $y = 2x + 1$. The equation becomes $6y^2 = y + 5$.Rearranging the terms,we get $6y^2 - y - 5 = 0$.To factorize,we look for two numbers that mu

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$0, -\frac{11}{12}$

Vedclass · Answer

(B) Let $y = 2x + 1$. The equation becomes $6y^2 = y + 5$.Rearranging the terms,we get $6y^2 - y - 5 = 0$.To factorize,we look for two numbers that multiply to $(6 	imes -5) = -30$ and add to $-1$.These numbers are $-6$ and $5$.So,$6y^2 - 6y + 5y - 5 = 0$.$6y(y - 1) + 5(y - 1) = 0$.$(6y + 5)(y - 1) = 0$.This gives $y = 1$ or $y = -\frac{5}{6}$.Substituting $y = 2x + 1$ back:Case $1$: $2x + 1 = 1 \implies 2x = 0 \implies x = 0$.Case $2$: $2x + 1 = -\frac{5}{6} \implies 2x = -\frac{5}{6} - 1 = -\frac{11}{6} \implies x = -\frac{11}{12}$.Thus,the solutions are $x = 0$ and $x = -\frac{11}{12}$.

Solve the following equation using the method of factorization: $6(2x+1)^2 = (2x+1) + 5$

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