Solve the following equation using the method | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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: $6(2x+1)^2 - (2x+1) - 5 = 0$

Similar Questions

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

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$

Solve the following equation using the method of factorization: $\frac{x-1}{x-2} + \frac{x-3}{x-4} = 3 \frac{1}{3}$ (where $x \neq 2, 4$).

Solve the following equation using the method of factorization and write its solution set: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$ $(x \neq 2, x \neq -3, x \neq 0)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module