Question

If $\sqrt{16x^2-48x+36}=2x+12$, then the sum of integral values of $x$ satisfying the equation is

• A. $9$
• B. $-9$
• C. $-8$
• D. $8$

Answer 1

The correct option is D $8$

$\sqrt{16x^2-48x+36}=2x+12$
$\Rightarrow \sqrt{4(4x^2-12x+9)}=2(x+6)$
$\Rightarrow 2\sqrt{(2x-3{)}^2}=2(x+6)$
$\Rightarrow |2x-3|=x+6(\because \sqrt{a^2}=|a|)$

Case $1:$ If $x\ge \frac{3}{2}$
then $2x-3=x+6$
$\Rightarrow x=9$

Case $2:$ If $x<\frac{3}{2}$
then $3-2x=x+6$
$\Rightarrow x=-1$

Sum $=9+(-1)=8$