Question

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

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

Answer 1

Answer: (B)

$b=4$

$2{\log }_{1/25}(bx+28)=-{\log }_5(12-4x-x^2)$
Let us simplify the term on the left hand side first,
$\Rightarrow {\log }_{1/5}(bx+28{)}^2=-{\log }_5(12-4x-x^2)$
$\Rightarrow {\log }_{1/5}(bx+28)=-{\log }_5(12-4x-x^2)$
$\Rightarrow -{\log }_5(bx+28)=-{\log }_5(12-4x-x^2)$
As both the terms are inside logarithm and to the same base, let us equate them.
$\Rightarrow bx+28=12-4x-x^2$
$\Rightarrow x^2+x(b+4)+16=0$ $\Rightarrow$ (1)
Since, the given equation have coincident roots.
Therefore equation (1) have equal roots.
${(b+4)}^2-64=0$
$\Rightarrow b=4,-12$
If $b=-12$, then $x$ will be $4$
But $x=4$ is not possible,
$\because 12-4x-x^2$ should be greater than $0$, but at $x=4,12-4x-x^2<0$
$\therefore b\ne -12$

Ans: B