Question

The sum of the possible value(s) of $a$ for which the equation
$2{\log }_{1/25}(ax+28)=-{\log }_5(12-4x-x^2)$ (wherever defined) has coincident roots, is

• A. $-8$
• B. $4$
• C. $16$
• D. $-12$

Answer 1

The correct option is A $-8$

$2{\log }_{1/25}(ax+28)=-{\log }_5(12-4x-x^2)$
$\Rightarrow 2{\log }_{5^{-2}}(ax+28)=-{\log }_5(12-4x-x^2)$
$\Rightarrow \frac{2}{-2}{\log }_5(ax+28)=-{\log }_5(12-4x-x^2)$
$\Rightarrow ax+28=12-4x-x^2$
$\Rightarrow x^2+(a+4)x+16=0$

The given equation has coincident roots.
So, $\Delta =0$
$\Rightarrow (a+4{)}^2-64=0$
$\Rightarrow a+4=\pm 8$
$\Rightarrow a=4,-12$
Required sum $=4-12=-8$