Let p , q ∈ R and (1-√3)200=2199(p+i q), t=√-1 Then p + q + q 2 and p - q + q 2 are roots of the equation.

Question

Let p , q ∈ R and (1-√3)200=2199(p+i q), t=√-1 Then p + q + q 2 and p - q + q 2 are roots of the equation. (A) x2-4 x+1=0 (B) x2+4 x+1=0 (C) x2-4 x-1=0 (D) x2+4 x-1=0

Tardigrade Admin · Accepted Answer

(1-√3 i )200=2199( p + iq ) 2200( cos (π/3)- i sin (π/3))200=2199( p + iq ) 2(-(1/2)- i (√3/2))= p + iq p =-1, q =-√3 α= p + q + q 2=2-√3 β= p - q + q 2=2+√3 α+β=4 α ⋅ β=1 equation x 2-4 x +1=0

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

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

$x^{2} + 4 x + 1 = 0$

$x^{2} - 4 x - 1 = 0$

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

$(1 - 3 i)^{200} = 2^{199} (p + i q)$
$2^{200} (cos \frac{π}{3} - i sin \frac{π}{3})^{200} = 2^{199} (p + i q)$
$2 (- \frac{1}{2} - i \frac{3}{2}) = p + i q$
$p = - 1, q = - 3$
$α = p + q + q^{2} = 2 - 3$
$β = p - q + q^{2} = 2 + 3$
$α + β = 4$
$α \cdot β = 1$
equation $x^{2} - 4 x + 1 = 0$

Q. Let p,q∈R and (1−3​)200=2199(p+iq),t=−1​ Then p+q+q2 and p−q+q2 are roots of the equation.

x2−4x+1=0

x2+4x+1=0

x2−4x−1=0

x2+4x−1=0