The set of all values of $m$ for which both the roots of the equation $x^{2}-(m+1) x+m+4=0$ are real and negative, is
(a) $(-\infty,-3] \cup[5, \infty)$
(b) $[-3,5]$
(c) $(-4,-3]$
(d) $(-3,-1]$
(c) $(-4,-3]$
The roots of the quadratic equation $x^{2}-(m+1) x+m+4=0$ will be real, if its discriminant is greater than or equal to zero.
$\therefore(m+1)^{2}-4(m+4) \geq 0$
$\Rightarrow(m-5)(m+3) \geq 0$
$\Rightarrow m \leq-3$ or $m \geq 5 \quad \ldots(1)$
It is also given that, the roots of $x^{2}-(m+1) x+m+4=0$ are negative.
So, the sum of the roots will be negative.
$\therefore$ Sum of the roots $<0$
$\Rightarrow m+1<0$
$\Rightarrow m<-1$ ...(2)
and product of zeros >0
$\Rightarrow m+4>0$
$\Rightarrow m>-4$ ....(3)
From (1), (2) and (3), we get,
$m \in(-4,-3]$
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.