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

Answer: (A)

$5$

Given $|\alpha -\beta {|}^2\le 16$

$\Rightarrow (\alpha +\beta {)}^2-4\alpha \beta \le 16$

$\Rightarrow 36-4b\le 16$

$\Rightarrow 4b\ge 20$

$\Rightarrow b\ge 5$

Least value of $b$ is $5$