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(A) Given equation: $6(2x+1)^2 - (2x+1) - 5 = 0$Let $m = 2x+1$.Substituting $m$ into the equation,we get: $6m^2 - m - 5 = 0$.Factorizing the quadratic

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given equation: $6(2x+1)^2 - (2x+1) - 5 = 0$Let $m = 2x+1$.Substituting $m$ into the equation,we get: $6m^2 - m - 5 = 0$.Factorizing the quadratic equation: $6m^2 - 6m + 5m - 5 = 0$.$6m(m-1) + 5(m-1) = 0$.$(6m+5)(m-1) = 0$.Therefore,$m = 1$ or $m = -\frac{5}{6}$.Case $1$: If $m = 1$,then $2x+1 = 1 \implies 2x = 0 \implies x = 0$.Case $2$: If $m = -\frac{5}{6}$,then $2x+1 = -\frac{5}{6} \implies 2x = -\frac{5}{6} - 1 \implies 2x = -\frac{11}{6} \implies x = -\frac{11}{12}$.Thus,the roots of the equation are $0$ and $-\frac{11}{12}$.

Solve the following equation using the method of factorization: $6(2x+1)^2 - (2x+1) - 5 = 0$

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