Quadratic Equations
Power of Complex Root — De Moivre's Theorem
nta_pyq_2023_apr
Grade 11
Question:
Let $\alpha,\beta$ be roots of $x^2-\sqrt{2}x+2=0$. Then $\alpha^{14}+\beta^{14}$ is equal to
$-64$
$-64\sqrt{2}$
$-128$
$-128\sqrt{2}$
Step-by-Step Solution
Key Concept: Roots: $\alpha=\sqrt{2}e^{i\pi/3}$, $\beta=\sqrt{2}e^{-i\pi/3}$. $\alpha^{14}+\beta^{14}=(\sqrt{2})^{14}\cdot2\cos(\frac{14\pi}{3})$.
$128\cdot(-1)=-128$.
Correct Answer: 3