Question

If the roots of the quadratic equation $x^2+6x+b=0$ are real and distinct and they differ by atmost $4$, then the least value of $b$ is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is A $5$

Roots of the quadratic equation $x^2+6x+b=0$ are real and distinct and they differ by atmost $4$.
i.e $B^2-4AC>0$ and $\frac{\sqrt{B^2-4AC}}{A}\le 4$
$B^2-4AC>0$
$\Rightarrow 36-4b>0$
$\Rightarrow b<9$
$\frac{\sqrt{B^2-4AC}}{A}\le 4$
$\Rightarrow \sqrt{36-4b}\le 4$
$\Rightarrow 36-4b\le 16$
$\therefore b\ge 5$
i.e Least value of $b$ is $5$.
Hence, option A.