Question

If $\alpha$ and ${\beta }^2$ are the roots of the equation $8x^2-10x+3=0,$ where ${\beta }^2>\frac{1}{2},$ then an equation whose roots are $(\alpha +i\beta {)}^{100}$ and $(\alpha -i\beta {)}^{100}$ is

• A. $x^2-x+1=0$
• B. $x^2+x+1=0$
• C. $x^2-x-1=0$
• D. $x^2+x-1=0$

Answer 1

The correct option is B $x^2+x+1=0$

$8x^2-10x+3=0$
$\Rightarrow (2x-1)(4x-3)=0$
Roots are $\alpha$ and ${\beta }^2$
So, $\alpha =\frac{1}{2}$ and
${\beta }^2=\frac{3}{4}(\because {\beta }^2>\frac{1}{2})$
$\Rightarrow \beta =\frac{\sqrt{3}}{2}$ or $-\frac{\sqrt{3}}{2}$

Now, given roots are $(\alpha +i\beta {)}^{100}$ and $(\alpha -i\beta {)}^{100}$
Roots are ${(\frac{1}{2}+i\frac{\sqrt{3}}{2})}^{100}$ and ${(\frac{1}{2}-i\frac{\sqrt{3}}{2})}^{100}$
or, ${(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})}^{100}$ and ${(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})}^{100}$

Sum of roots $=2Rez=2\cos \frac{100\pi }{3}=2\cos (33\pi +\frac{\pi }{3})=-2\cos \frac{\pi }{3}=-1$
Product of roots $=1$

So, required quadratic equation is $x^2+x+1=0$