Question

For the equation $x^2-(k+1)x+(k^2+k-8)=0$ if one root is greater then $2$ and other is less than $2$, then $k$ lies between

• A. $-2$ & $3$
• B. $2$ & $-2$
• C. $2$ & $-3$
• D. None of these

Answer 1

Answer: (A)

$-2$ & $3$

For the equation $x^2-(k+1)x+(k^2+k-8)=0$, one root is greater than $2$ and other is less than $2$.
i.e as $a>0$, $f\left(2\right)<0$
$\Rightarrow 4-2k-2+k^2+k-8<0$
$\Rightarrow k^2-k-6<0$
$\therefore k\in (-2,3)$
Hence, option A.