Question

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]

Answer 1

(c) $m\in (-4,-3]$

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.

$$\begin{gathered} \therefore {\left(m+1\right)}^2-4\left(m+4\right)\ge 0 \\ \Rightarrow \left(m-5\right)\left(m+3\right)\ge 0 \\ \Rightarrow m\le -3\mathrm{or}m\ge 5...\left(1\right) \end{gathered}$$

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

$$\begin{gathered} \Rightarrow m+1<0 \\ \Rightarrow m<-1...\left(2\right) \end{gathered}$$

and product of zeros >0

$$\begin{gathered} \Rightarrow m+4>0 \\ \Rightarrow m>-4...\left(3\right) \end{gathered}$$

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.