Question

Find $k$ for which one root of the equation $x^2-(k+1)x+k^2+k-8=0$ is greater than 2 and other is less than 2

• A. $k\in [-2,3]$
• B. $k\in (-2,\infty )$
• C. $k\in (-2,3)$
• D. $k\in (-\infty ,-2)\cup (3,\infty )$

Answer 1

The correct option is C $k\in (-2,3)$

Here a >0 for given expression $f\left(x\right)=x^2-(k+1)x+(k^2+k-8)$.
For roots lying on either sides of 2, we must have
$\begin{array}{cc}& \Rightarrow af\left(d\right)<0\\ & \Rightarrow \left(1\right)f\left(2\right)<0\\ \Rightarrow & 4-(k+1)2+(k^2+k-8)<0\\ \Rightarrow & k^2-3k+2k-6<0\\ \Rightarrow & k(k-2)+2(k-3)<0\\ \Rightarrow & (k+2)(k-3)<0\\ \Rightarrow & k\in (-2,3)\end{array}$