Question

If $a{x}^2-bx+5=0$ does not have two distinct real roots, then the minimum value of $5a+b$ is

• A. $-1$
• B. $0$
• C. $-2$
• D. $-5$

Answer 1

Answer: (A)

$-1$

$f\left(x\right)=a{x}^2-bx+5=0$
$\because f\left(x\right)$ does not have two distinct real roots .
either $f\left(x\right)\ge 0\forall x\in R$ or $f\left(x\right)\le 0\forall x\in R$
But, $f\left(0\right)=5>0$, so $f\left(x\right)\ge 0\forall x\in R$
Also,
$f(-5)\ge 0\Rightarrow 25a+5b+5\ge 0$
$\Rightarrow 5a+b\ge -1$
$\therefore$ Least value of $5a+b$ is $-1$ .