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 is lies between -2 and 3.

Answer 1

If $\alpha ,\beta$ be the roots of the given equation f (x) = 0 then the given condition
$\alpha <2<beta$ ....(1)
Also f(x) = 1 . $(x-\alpha )(x-\beta ),\alpha <\beta$ ...(2)
We conclude from (1) that the roots of the given equation must be real as order relation does not exist in complex numbers. Secondly from(2) we conclude that f (x) = -ive for all values of x which
lie between $\alpha$ and $\beta$. $\therefore f\left(2\right)=-ive$
$\therefore \delta >0$ distinct f(2) = -ive
$\therefore (k+1{)}^2-4(K^2+k-8)=+ive$
or $k^2+2k-33=$ $-ive$
or ($3k+11\left)\right(k-3)$ = $-ive$
$\therefore -\frac{11}{3}<k<3$ ....(3)
Again $f\left(2\right)=-ive$
$\Rightarrow 4-2(k+1)+k^2+k-8$ is $-ive$
$k^2-k-6$ = $-ive$ or $(k+2)(k-3)$ = $-ive$
$\therefore$ $-2<k<3$ ...(4)
Hence k should be so chosen as to satisfy both (3) and (4). In other words it will satisfy the common region or intersection of the intervals given by (3) and (4).