Question

For the equation, $x^2-(k+1)x+(k^2+k-8)=0$ if one root is greater than $2$ and other is less than $2$, then prove that k lies between $-2$ and $3$.

Answer 1

We are given that equation

$x^2-(k+1)x+(k^2+k-8)=0$.

$a=1$ which is greater than zero.

let, roots are $\alpha$ & $\beta$ & $\alpha <2$ & $\beta >2$

here function will be negative at $x=2$

$\therefore f\left(x\right)=x^2-(k+1)x+(k^2+k-8)=0$

$\therefore f\left(2\right)=(2{)}^2-(k+1\left)\right(2)+(k^2+k-8)$

$=4-(2k+2)+(k^2+k-8)$

$=4-2k-2+k^2+k-8$

$f\left(2\right)=k^2-k-6$

But $f\left(2\right)<0$.

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

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

Only for $k\in (-2,3)$, function will be negative .