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Difference between revisions of "1953 AHSME Problems/Problem 3"
Line 7: Line 7:
 
Trying each case out, we see  
 
Trying each case out, we see  
 
<math>(x+iy)(x-iy)=x^2+xyi-xyi+(iy)(-iy)=x^2+(-1)(-y^2)=x^2+y^2</math>
 
<math>(x+iy)(x-iy)=x^2+xyi-xyi+(iy)(-iy)=x^2+(-1)(-y^2)=x^2+y^2</math>
 +
 
So <math>\boxed{\text{D}}</math> works
 
So <math>\boxed{\text{D}}</math> works
  
 
~mathsolver101
 
~mathsolver101

Revision as of 14:19, 31 July 2015

The factors of the expression $x^2+y^2$ are:

$\textbf{(A)}\ (x+y)(x-y) \qquad \textbf{(B)}\ (x+y)^2 \qquad \textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\\ \textbf{(D)}\ (x+iy)(x-iy)\qquad \textbf{(E)}\ \text{none of these}$

Solution

Trying each case out, we see $(x+iy)(x-iy)=x^2+xyi-xyi+(iy)(-iy)=x^2+(-1)(-y^2)=x^2+y^2$

So $\boxed{\text{D}}$ works

~mathsolver101

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