Question

If the quadratic equation $z^2+(2+4i)z+(c+5i)=0$ where $c$ is non zero real number, has real roots then the value of $c$ equals

• A. $\frac{16}{15}$
• B. $\frac{65}{16}$
• C. $\frac{15}{16}$
• D. None of these

Answer 1

Answer: (C)

$\frac{15}{16}$

$z^2+(2+4i)z+c+5i=0$ ...(1)
Let $a$ be the root of the equation (1).
Therefore, $a^2+(2+4i)a+c+5i=0$ ...(2)
Since, a is real.
Therefore, $a=\overline{a}$
Taking conjugate of equation (2) we get,
${\overline{a}}^2+(2-4i)\overline{a}+c-5i=0$
$\Rightarrow a^2+(2-4i)a+c-5i=0$ ...(3)
Substracting (2) & (3) we get,
$8ia+10i=0\Rightarrow a=\frac{-5}{4}$
Substituting $a=\frac{-5}{4}$ in equation (2) we get,
$\Rightarrow {\left(\frac{-5}{4}\right)}^2-(2-4i)\frac{5}{4}+c-5i=0$
$\therefore c=\frac{15}{16}$
Ans: C