The set of all values of m for which both the roots of the equation x

Solution:

(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.