Question

If $\alpha$ is a complex constant such that $\alpha z^2+z+\overline{\alpha }=0$ has a real root then

• A. $\alpha +\overline{\alpha }=1$
• B. $\alpha +\overline{\alpha }=0$
• C. $\alpha +\overline{\alpha }=-1$
• D. The absolute value of the real root is $1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\alpha +\overline{\alpha }=1$
C $\alpha +\overline{\alpha }=-1$
D The absolute value of the real root is $1$

The given equation is $\alpha z^2+z+\overline{\alpha }=0$

Let $x$ be a real root of the given equation.

Then, $\overline{x}=x$

Since $x$ is a root of the given equation thus we have-

$\alpha x^2+x+\overline{\alpha }=0$ ....................................$\left(i\right)$

Applying conjugate on both sides and using the fact that $\overline{x}=x$ we get-

$\overline{\alpha x^2+x+\overline{\alpha }}=\overline{0}$

$\Rightarrow \overline{\alpha }.{\overline{x}}^2+\overline{x}+\overline{\overline{\alpha }}=0$

{using the fact that $\overline{z+w}=\overline{z}+\overline{w}$ and $\overline{zw}=\left(\overline{z}\right)\left(\overline{w}\right)$ and conjugate of $\overline{z}$ is $z$}

$\Rightarrow \overline{\alpha }{x}^2+x+\alpha =0$ .....................................$\left(ii\right)$

Subtracting equation $\left(ii\right)$ from $\left(i\right)$ we get-

$(\alpha x^2+x+\overline{\alpha })-(\overline{\alpha }{x}^2+x+\alpha )=0$

$\Rightarrow (\alpha -\overline{\alpha })x^2+(\overline{\alpha }-\alpha )=0$

$\Rightarrow (\alpha -\overline{\alpha })x^2=-(\overline{\alpha }-\alpha )=\alpha -\overline{\alpha }$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

Hence, $\left|x\right|=1$, i.e.- the absolute value of the real root is 1.

Again, multiplying equation $\left(i\right)$ by $\overline{\alpha }$ and multiplying equation $\left(ii\right)$ by $\alpha$ and subtracting both we get-

$\overline{\alpha }(\alpha x^2+x+\overline{\alpha })-\alpha (\overline{\alpha }{x}^2+x+\alpha )=0$

$\Rightarrow \overline{\alpha }\alpha x^2+\overline{\alpha }x+\overline{\alpha }\overline{\alpha }-(\alpha \overline{\alpha }{x}^2+\alpha x+\alpha \alpha )=0$

$\Rightarrow \overline{\alpha }x+{\overline{\alpha }}^2-\alpha x-{\alpha }^2=0$

$\Rightarrow \overline{\alpha }x-\alpha x={\alpha }^2-{\overline{\alpha }}^2=(\alpha +\overline{\alpha })(\alpha -\overline{\alpha })$

$\Rightarrow (\overline{\alpha }-\alpha )x=(\alpha +\overline{\alpha })(\alpha -\overline{\alpha })$

$\Rightarrow -(\alpha -\overline{\alpha })x=(\alpha +\overline{\alpha })(\alpha -\overline{\alpha })$

$\Rightarrow x=-(\alpha +\overline{\alpha })$

$\Rightarrow -(\alpha +\overline{\alpha })=\pm 1$ {Since $x=\pm 1$}

$\Rightarrow (\alpha +\overline{\alpha })=\mp 1$

Hence, $(\alpha +\overline{\alpha })=1$ or $(\alpha +\overline{\alpha })=-1$.

Hence, the correct answers are-

$(\alpha +\overline{\alpha })=1$

$(\alpha +\overline{\alpha })=-1$

The absolute value of the real root is 1.