1972 AHSME Problems/Problem 3: Difference between revisions

Revision as of 00:54, 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 $\boxed{\textbf{(C) }-1}.$ ~lopkiloinm

Revision as of 00:53, 7 November 2020 view source Lopkiloinm (talk \| contribs) editor 496 edits →Solution ← Older edit		Revision as of 00:54, 7 November 2020 view source Lopkiloinm (talk \| contribs) editor 496 edits →Solution Newer edit →
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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 <cmath>\boxed{\textbf{(C) }-1.</cmath> ~lopkiloinm		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 <cmath>\boxed{\textbf{(C) }-1}.</cmath> ~lopkiloinm

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1972 AHSME Problems/Problem 3: Difference between revisions

Revision as of 00:54, 7 November 2020

Problem 3

Solution