Question

Given the cubic equation $x ^3-2 kx ^2-4 kx + k ^2=0$. If one root of the equation is less than 1 , other root is in the interval $(1,4)$ and the $3^{\text {rd }}$ root is greater than 4 , then the value of $k$ lies in the interval $(a+\sqrt{b}, b(a+\sqrt{b}))$ where $a, b \in N$. Find the value of $(a+b)^3+(a b+2)^2$.

Answer 1

$ f(x)=x^3-2 k x^2-4 k x+k^2=0 $
$\text { note that } f(0)=k^2>0$
$\Rightarrow f(1)>0 $
$\Rightarrow 1-2 k-4 k+k^2>0 $
$k^2-6 k+1>0 $
${[k-(3+2 \sqrt{2})][k-(3-2 \sqrt{2})]>0 }$

$\text { Also } f (4)<0 $
$\Rightarrow 64-32 k -16 k + k ^2<0$
$ k ^2-48 k +64<0 $
$ ( k -24)^2<512$
$ ( k -24+16 \sqrt{2})( k -24-16 \sqrt{2})<0 $
$ {[ k -8(3-2 \sqrt{2})][ k -8(3+2 \sqrt{2})]<0}$

$(1) \cap(2) \Rightarrow 3+2 \sqrt{2}< k <8(3+2 \sqrt{2}) $
$3+\sqrt{8}< k <8(3+\sqrt{8}) $
$\therefore a =3 ; b =8$
$( a + b )^3+( ab +2)^2 \Rightarrow (11)^3+(26)^2=1331+676=2007$