Difference between revisions of "1972 AHSME Problems/Problem 3"

(Created page with "== Problem 3 == If <math>x=\dfrac{1-i\sqrt{3}}{2}</math> where <math>i=\sqrt{-1}</math>, then <math>\dfrac{1}{x^2-x}</math> is equal to <math>\textbf{(A) }-2\qquad \textbf{...")
 
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== Solution ==
 
== Solution ==
  
Using DeMoivre's theorem, we can calculate <math>x^2=\frac{1+i\sqrt{3}}{2}</math> The denominator is therefore <math>-1</math> which makes the answer <math>textbf{(B) }-1</math>. ~lopkiloinm
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Using DeMoivre's theorem, we can calculate <math>x^2=\frac{1+i\sqrt{3}}{2}</math> The denominator is therefore <math>-1</math> which makes the answer <math>\boxed{\textbf{(C) }-1</math>. ~lopkiloinm

Revision as of 01:53, 7 November 2020

Problem 3

If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$, then $\dfrac{1}{x^2-x}$ is equal to

$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad  \textbf{(E) }2$

Solution

Using DeMoivre's theorem, we can calculate $x^2=\frac{1+i\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer $\boxed{\textbf{(C) }-1$ (Error compiling LaTeX. Unknown error_msg). ~lopkiloinm