Question

If $z_0=\alpha +i\beta ,i=\sqrt{-1}$ then the roots of the cubic equation $x^3-2(1+\alpha )x^2+(4\alpha +{\alpha }^2+{\beta }^2)x-2({\alpha }^2+{\beta }^2)=0$ are

• A. $2,z_0,{\overline{z}}_0$
• B. $1,z_0,-z_0$
• C. $2,z_0,-{\overline{z}}_0$
• D. $2,-z_0,{\overline{z}}_0$

Answer 1

The correct option is A $2,z_0,{\overline{z}}_0$

Let $f\left(x\right)=x^3-2(1+\alpha )x^2+(4\alpha +{\alpha }^2+{\beta }^2)x-2({\alpha }^2+{\beta }^2)$
Then
$f\left(2\right)=8-8-8\alpha +8\alpha +2{\alpha }^2+2{\beta }^2-2{\alpha }^2-2{\beta }^2=0$
Hence $x=2$ is one root
Now factorizing we get
$(x-2)(x^2-2\alpha x+{\alpha }^2+{\beta }^2)=0$
And for
$x^2-2\alpha x+{\alpha }^2+{\beta }^2=0$
$\Rightarrow x=z_0,\overline{z_0}$
Therefore,
$2,z_0,\overline{z_0}$ are roots