Question

Consider a quadratic equation $a{z}^2+bz+c=0$, where a, b, c are complex numbers, then the condition that equation has one purely real root is,

• A. $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}+c\overline{b}\left)\right(a\overline{b}-\overline{a}b)$
• B. $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}-c\overline{b}\left)\right(a\overline{b}+\overline{a}b)$
• C. $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}+c\overline{b}\left)\right(a\overline{b}+\overline{a}b)$
• D. $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}-c\overline{b}\left)\right(a\overline{b}-\overline{a}b)$

Answer 1

The correct option is D $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}-c\overline{b}\left)\right(a\overline{b}-\overline{a}b)$

Let $z_1$ (purely real) be a root of the given equation. Then,

$z_1={\overline{z}}_1$

and $a{z}_1^2+b{z}_1+c=0$ .....(1)

or $a{z}_1^2+b{z}_1+c=\overline{0}$

or $\overline{a}{\overline{z}}_1^2+\overline{b}{\overline{z}}_1+\overline{c}=0$

or $\overline{a}{z}_1^2+\overline{b}{z}_1+\overline{c}=0$.....(2)

Now (1) and (2) must have one common root. Hence, $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}-c\overline{b}\left)\right(a\overline{b}-\overline{a}b)$