Question

Given the cubic equation $x^3-2k{x}^2-4kx+k^2=0$. 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 $(a+b{)}^3+(ab+2{)}^2$.

Answer 1

Answer: 2007

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

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

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