Difference between revisions of "1953 AHSME Problems/Problem 3"

 
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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>
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So <math>\boxed{\text{D}}</math> works
 
So <math>\boxed{\text{D}}</math> works
  
~mathsolver101
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==See Also==
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{{AHSME 50p box|year=1953|num-b=2|num-a=4}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:17, 1 April 2017

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

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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