Question

If the roots of the quadratic equation $x^2-4x-{\log }_{3}a=0$ are real, then the least value of $a$ is

• A. $81$
• B. $\frac{1}{81}$
• C. $\frac{1}{64}$
• D. $9$

Answer 1

The correct option is B $\frac{1}{81}$

Given roots are real
$\Rightarrow b^2-4ac\ge 0$
$\Rightarrow 16+4\left({\log }_{3}a\right)\ge 0$
$lo{g}_{3}a\ge -4$
$a\ge 3^{-4}$
Least value of $a$ is $\frac{1}{3^4}=\frac{1}{81}$ .