Question

Find the set of values of $m$ for which exactly one root of the equation
$x^2+mx+(m^2+6m)=0$ lie on $(-2,0)$

Answer 1

Let,

$f\left(x\right)=x^2+mx+(m^2+6m)$

As per question given that exactly one root of the equation lies between -2 and 0.

Hence, only one is negative.

Therefore $f(-2)\times f\left(0\right)<0$

Putting the value in the function ,we get

$[{(-2)}^2-2m+m^2+6m]\times [m^2+6m]<0$

$\Rightarrow [m^2+4m+4]\times m(m+6)<0$

$\Rightarrow {(m+2)}^2\times m\times (m+6)<0$

Hence, the range of m becomes

$m\in (-6,0)-\{-2,\}$

We can write it as

$m\in (-6,-2)\cup (-2,0)$

Hence, this is the answer .