Question

The quadratic equation whose roots are $\frac{\sqrt{2}+\sqrt{3}i}{\sqrt{2}-\sqrt{3}i}$ and $\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}+\sqrt{3}i}$ is
(Note : $i=\sqrt{-1}$)

• A. $5x^2-2x+5=0$
• B. $5x^2+2x+5=0$
• C. $5x^2+2x-5=0$
• D. $5x^2-2x-5=0$

Answer 1

The correct option is B $5x^2+2x+5=0$

Let the equation be $a{x}^2+bx+c=0$

The sum of roots will be $=\frac{-b}{a}$
$=\frac{\sqrt{2}+\sqrt{3}i}{\sqrt{2}-\sqrt{3}i}+\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}+\sqrt{3}i}$
$=\frac{(\sqrt{2}+\sqrt{3}i{)}^2+(\sqrt{2}-\sqrt{3}i{)}^2}{{\sqrt{2}}^2+{\sqrt{3}}^2}$
$=\frac{2-3+2\sqrt{6}i+2-3-2\sqrt{6}i}{5}$
$=\frac{4-6}{5}$
$=\frac{-2}{5}$

Hence
$\frac{-b}{a}=\frac{-2}{5}$ Or $\frac{b}{a}=\frac{2}{5}$ ...(i)
Similarly product of roots will be
$\frac{c}{a}=\frac{\sqrt{2}+\sqrt{3}i}{\sqrt{2}-\sqrt{3}i}\times \frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}+\sqrt{3}i}$

Hence
$\frac{c}{a}=1$ ...(ii)
Now
$a{x}^2+bx+c=0$ can be written as
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
from (i) and (ii)
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
$\Rightarrow x^2+\frac{2}{5}x+1=0$
$5x^2+2x+5=0$
Hence the required equation is
$5x^2+2x+5=0$.