Question

If $A=\{a\in R|\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\}$ has atleast one real root. Then the value of $\frac{\sum a^2}{2}$ is

Answer 1

$A\equiv (1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0$
Let the real root of equation be $\alpha$. Then,
$(1+2i){\alpha }^3-2(3+i){\alpha }^2+(5-4i)\alpha +2a^2=0$
Equating imaginary part zero, we get
$2{\alpha }^3-2{\alpha }^2-4\alpha =0$
$\Rightarrow \alpha ({\alpha }^2-\alpha -2)=0$
$\Rightarrow \alpha =0$ or $\alpha =-1,2$

Now, equating real part to zero, we have
${\alpha }^3-6{\alpha }^2+5\alpha +2a^2=0$
$\alpha =0\Rightarrow a=0$
$\alpha =-1\Rightarrow a=\pm \sqrt{6}$
$\alpha =2\Rightarrow a=\pm \sqrt{3}$
$\therefore \sum a^2=0^2+(\sqrt{6}{)}^2+(-\sqrt{6}{)}^2+(\sqrt{3}{)}^2+(-\sqrt{3}{)}^2$
$=18$

$\Rightarrow \frac{\sum a^2}{2}=9$