Difference between revisions of "1990 AHSME Problems/Problem 22"

(Created page with "== Problem == If the six solutions of <math>x^6=-64</math> are written in the form <math>a+bi</math>, where <math>a</math> and <math>b</math> are real, then the product of those...")
 
 
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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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This equation is <math>r^6e^{6\theta i}=2^6e^{(\pi\pm 2k\pi) i}</math>. Solving in the usual way, <math>r=2</math> and <math>\theta\in\{\pm 30^\circ,\pm 90^\circ,\pm 150^\circ\}</math>.
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Thus there are only two solutions with positive real part, and they are conjugates, so their product is <math>r^2=4</math> which is <math>\fbox{D}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:17, 4 February 2016

Problem

If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is


$\text{(A) } -2\quad \text{(B) } 0\quad \text{(C) } 2i\quad \text{(D) } 4\quad \text{(E) } 16$

Solution

This equation is $r^6e^{6\theta i}=2^6e^{(\pi\pm 2k\pi) i}$. Solving in the usual way, $r=2$ and $\theta\in\{\pm 30^\circ,\pm 90^\circ,\pm 150^\circ\}$.

Thus there are only two solutions with positive real part, and they are conjugates, so their product is $r^2=4$ which is $\fbox{D}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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