Question

If the roots of the quadratic equation $x^2-4x-{\log }_3\left(a\right)=0$ are real, then the least value of $a$ is

• A. $81$
• B. $\frac{1}{81}$
• C. $\frac{1}{64}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{81}$

Step 1: Solve for the discriminant value

Given that the roots of the quadratic equation $x^2-4x-{\log }_3\left(a\right)=0$ are real $\Rightarrow$Discriminant $D\ge 0$

We know that for a quadratic equation $a{x}^2+bx+c=0$, Discriminant $D=b^2-4ac$

Here,

$\begin{array}{cccl}& D& =& {\left(-4\right)}^2-4\left(1\right)\left(-{\log }_3\left(a\right)\right)\\ \Rightarrow & D& =& 16+4{\log }_3\left(a\right)\end{array}$

Step 2: Solve for the value of $a$

Given that $D\ge 0$

$\begin{array}{crclc}\Rightarrow & 16+4{\log }_3\left(a\right)& \ge & 0& \\ \Rightarrow & 4{\log }_3\left(a\right)& \ge & -16& \\ \Rightarrow & {\log }_3\left(a\right)& \ge & -4& \\ \Rightarrow & a& \ge & 3^{-4}& \left[\because \forall {\log }_m\left(x\right)=y\Rightarrow x=m^y\right]\\ \Rightarrow & a& \ge & \frac{1}{81}& \end{array}$

Thus the minimum value of $a$ is $\frac{1}{81}$.

Hence, the correct option is option (B) i.e. $\frac{1}{81}$.