Question

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

• A. $(0,6)-\left\{2\right\}$
• B. $(-6,0)-\{-2\}$
• C. $(-6,0)$
• D. $(-\infty ,-6)\cup (0,\infty )$

Answer 1

The correct option is B $(-6,0)-\{-2\}$

Step-1: Making leading coefficient positive.
Let $f\left(x\right)=x^2+mx+m^2+6m$
Step-2:
Case-l: When no root is - 2 or 0
$f(-2)f\left(0\right)<0$
$\Rightarrow (4-2m+m^2+6m)(m^2+6m)<0\Rightarrow (m^2+4m+4)(m^2+6m)<0$
$\Rightarrow (m+2{)}^{2}m(m+6)<0\Rightarrow m\in (-6,0)-\{-2\}$
Case-II: When one of the root is $-2$ or 0
(i) if $f(-2)=0$
$\Rightarrow 4-2m+m^2+6m=0$
$\Rightarrow m^2+4m+4=0\Rightarrow m=-2$
For $m=-2$ equation: $x^2-2x-8=0\Rightarrow x=4,-2\Rightarrow$ No root in $(-2,0)$
(ii) If $f\left(0\right)=0$
$\Rightarrow m^2+6m=0\Rightarrow m=0,-6$
For $m=0$ equation: $x^2=0\Rightarrow x=0,0\Rightarrow$ No root in $(-2,0)$
Hence, $m\in (-6,0)-\{-2\}$