Question

If $a{x}^2+bx+8=0$ has no distinct real roots, then least value of $(4a+b)$ is (where $a$ and $b$ are real)

• A. $-3$
• B. $-2$
• C. $-4$
• D. None of these

Answer 1

Answer: (B)

$-2$

$a{x}^2+bx+8=0$
$D<0$
$b^2-32a<0$
$8\left(4a\right)-b^2\ge 0$
$\Rightarrow 8\left(4a\right)\ge b^2$
$\Rightarrow 4a\ge \frac{b^2}{8}$
$\Rightarrow 4a+b\ge \frac{b^2}{8}+b$
$forminimina$
$Let$
$y=\frac{b^2}{8}+b$
$\frac{dy}{db}=\frac{2b}{8}+1$
$\frac{dy}{db}=0$
$\frac{2b}{8}+1=0$
$\Rightarrow \frac{b}{4}=-1$
$\Rightarrow b=-4$
$\frac{{(-4)}^2}{8}-4$
$=-2$
$hence$
$4a+b\ge -2$