Question

Solve the following quadratic equations by factorization method:
$3(x-4{)}^2-5(x-4)=12$

• A. $\{7,\frac{8}{3}\}$
• B. $\{-7,\frac{8}{3}\}$
• C. $\{7,\frac{-8}{3}\}$
• D. None of these

Answer 1

The correct option is A $\{7,\frac{8}{3}\}$

$3(x-4{)}^2-5(x-4)=12$

$\Rightarrow$ $3(x^2-8x+16)-5x+20=12$

$\Rightarrow$ $3x^2-24x+48-5x+20=12$

$\Rightarrow$ $3x^2-29x+56=0$

$\Rightarrow$ $3x^2-21x-8x+56=0$

$\Rightarrow$ $3x(x-7)-8(x-7)=0$

$\Rightarrow$ $(x-7)(3x-8)=0$

$\Rightarrow$ $x-7=0$ and $3x-8=0$

$\therefore$ $x=7$ and $x=\frac{8}{3}$