Question

Let $p , q \in R$ and $(1-\sqrt{3})^{200}=2^{199}(p+i q), t=\sqrt{-1}$ Then $p + q + q ^2$ and $p - q + q ^2$ are roots of the equation

• A. $x^2-4 x+1=0$
• B. $x^2+4 x+1=0$
• C. $x^2-4 x-1=0$
• D. $x^2+4 x-1=0$

Answer 1

The Correct Option is A

The correct answer is (A) : $x^2-4 x+1=0$
$(1-\sqrt{3}i{)}^{200}=2^{199}(p+iq)$
$2^{200}{\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)}^{200}=2^{199}(p+iq)$
$2\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=p+iq$
$p=-1,q=-\sqrt{3}$
$\alpha =p+q+q^2=2-\sqrt{3}$
$\beta =p-q+q^2=2+\sqrt{3}$
$\alpha +\beta =4$
$\alpha \cdot \beta =1$
equation $x^2-4x+1=0$