Question

Find the range of valaues of $k$ 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. $k\in (-\infty ,-2)\cup (3,\infty )$
• D. $k\in [3,\infty )$

Answer 1

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

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
$\begin{array}{cc}& \Rightarrow a\cdot f\left(2\right)<0\\ & \Rightarrow \left(1\right)f\left(2\right)<0\\ \Rightarrow & 4-(k+1)2+(k^2+k-8)<0\\ \Rightarrow & k^2-k-6<0\\ \Rightarrow & k^2-3k+2k-6<0\\ \Rightarrow & k(k-3)+2(k-3)<0\\ \Rightarrow & (k+2)(k-3)<0\\ \Rightarrow & k\in (-2,3)\end{array}$