Question

The set of values of '$m$' for which both roots of the equation $x^2+(m+1)x+m+4=0$ are real and negative is

• A. $-3<m\le -1$
• B. $m\ge 5$
• C. $-3<m\le 5$
• D. $-3\ge m$ or $m\ge 5$

Answer 1

Answer: (B)

$m\ge 5$

We have, $x^2+(m+1)x+m+4=0$
Discriminant is $\Delta =(m+1{)}^2-4(m+4)=m^2-2m-15=(m+3\left)\right(m-5)$
For roots to be real $\Delta \ge 0\Rightarrow m\in (-\infty ,-3)\cup [5,\infty )\Rightarrow 1$
Also for root to be negative $\alpha \beta >0\text{\&}\alpha +\beta <0$
$\Rightarrow m+4>0\text{\&}-(m+1)<0\Rightarrow m>-4\text{\&}m>-1$
$\Rightarrow m>-1\Rightarrow \left(2\right)$
Thus from (1) and (2) we have $m\ge 5$
Hence option 'B' is correct choice.