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

Options

• \[( - \infty , - 3] \cup [5, \infty )\]

• [−3, 5]

• (−4, −3]

•  (−3, −1]

Answer 1

\[m \in ( - 4, - 3]\] 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.

\[\therefore \left( m + 1 \right)^2 - 4\left( m + 4 \right) \geq 0\]

\[ \Rightarrow \left( m - 5 \right)\left( m + 3 \right) \geq 0\]

\[ \Rightarrow m \leq - 3 \text { or } m \geq 5 . . . (1)\]

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.