Question

The value(s) of 'b' for which the equation, $2{\log }_{1/25}(bx+28)=-{\log }_5\left(12-4x-x^2\right)$ has coincident roots, is/are -

• A. $b=-12$
• B. $b=4$
• C. $b=4$ or $b=-12$
• D. $b=-4$ or $b=12$

Answer 1

Answer: (B)

${\log }_{\frac{1}{5}}(bx+28)={\log }_{\frac{1}{5}}\left(12-4x-x^2\right)$
$\Rightarrow bx+28=12-4x-x^2$
$x^2+(b+4)x+16=0$
$\Rightarrow \text{ since }D=0(b+4{)}^2-64=0$
$\Rightarrow b=4,-12$
$\text{ At }b=-12,x=4$
Put in log domain $\Rightarrow$ No solution
at $b=4$ satisfies the domain