Question

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

• A. None of the above
• B. $(3,\infty )$
• C. $(-2,3)$
• D. $(-\infty ,-2)$

Answer 1

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

Let $f\left(x\right)=x^2-(k+1)x+k^2+k-8$


Here, $f\left(2\right)<0$

$\Rightarrow (2{)}^2-(k+1)2+k^2+k-8<0$

$\Rightarrow 4-2k-2+k^2+k-8<0$

$\Rightarrow k^2-k-6<0$

$\Rightarrow (k+2)(k-3)<0$

$\Rightarrow k\in (-2,3)$