Question

The roots of a quadratic equation $x^2-4x-a=0$ are real. Then what is the least value of $a$ ?

• A. 64
• B. -4
• C. $\frac{1}{64}$
• D. 4

Answer 1

The correct option is B

-4

We have, $x^2-4x-a=0$
If roots are real, then $D\ge 0$
$\therefore 16+4a\ge 0$
$\Rightarrow 4a\ge -16\Rightarrow a\ge -4$