Question

Solve the following quadratic equations by factorization method:
$3(x^2-6)=x(x+7)-3$

• A. $\{-\frac{3}{2},5\}$
• B. $\{\frac{3}{2},5\}$
• C. $\{-\frac{3}{2},-5\}$
• D. None of these

Answer 1

The correct option is A $\{-\frac{3}{2},5\}$

$3(x^2-6)=x(x+7)-3$

$\Rightarrow$ $3x^2-18=x^2+7x-3$

$\Rightarrow$ $3x^2-18-x^2-7x+3=0$

$\Rightarrow$ $2x^2-7x-15=0$

$\Rightarrow$ $2x^2-10x+3x-15=0$

$\Rightarrow$ $2x(x-5)+3(x-5)=0$

$\Rightarrow$ $(x-5)(2x+3)=0$

$\Rightarrow$ $x-5=0$ and $2x+3=0$

$\therefore$ $x=5$ and $x=-\frac{3}{2}$