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

Answer: (A)

$x^2-4 x+1=0$

Answer 2

The correct answer is (A) : $x^2-4 x+1=0$
(1−3​i)200=2199(p+iq)
2200(cos3π​−isin3π​)200=2199(p+iq)
2(−21​−i23​​)=p+iq
p=−1,q=−3​
α=p+q+q2=2−3​
β=p−q+q2=2+3​
α+β=4
α⋅β=1
equation x2−4x+1=0