Question

Given $l{x}^2-mx+5=0$ does not have distinct real roots then minimum value of $5l+m$ is

• A. $5$
• B. $-5$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

Given that

$l{x}^2-mx+5=0$ does not have $2$ distinct real roots, the minimum value of $5l+m$

For $f\left(x\right)=l{x}^2-mx+5$

$\therefore l{x}^2-mx+5=0$ does not have distinct real root,

$f\left(x\right)\ge 0$ $\forall x\in R$ and $f\left(x\right)\le 0\forall x\in R$

$\therefore f\left(0\right)=l{x}^2-mx+5$

$=l(0{)}^2-m(0)+5$

$=5$

$f\left(0\right)=5>0$

$\therefore f\left(x\right)\ge 0\forall x\in R$

$f(-5)\ge 0=l(-5{)}^2-m(-5)+5$

$\Rightarrow 25l+5m+5\ge 0$

$\Rightarrow 5(5l+m+1)\ge 0$

$\Rightarrow 5l+m\ge -1$

$\therefore$ Min. value of $5l+m$ is $-1$