Difference between revisions of "1958 AHSME Problems/Problem 9"

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== Problem ==
 
== Problem ==
A value of <math> x</math> satisfying the equation <math> x^2 \plus{} b^2 \equal{} (a \minus{} x)^2</math> is:
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A value of <math> x</math> satisfying the equation <math> x^2 + b^2 = (a - x)^2</math> is:
  
<math> \textbf{(A)}\ \frac{b^2 \plus{} a^2}{2a}\qquad  
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<math> \textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad  
\textbf{(B)}\ \frac{b^2 \minus{} a^2}{2a}\qquad  
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\textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad  
\textbf{(C)}\ \frac{a^2 \minus{} b^2}{2a}\qquad  
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\textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad  
\textbf{(D)}\ \frac{a \minus{} b}{2}\qquad  
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\textbf{(D)}\ \frac{a - b}{2}\qquad  
\textbf{(E)}\ \frac{a^2 \minus{} b^2}{2}</math>
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\textbf{(E)}\ \frac{a^2 - b^2}{2}</math>
  
 
== Solution ==
 
== Solution ==

Latest revision as of 23:06, 13 March 2015

Problem

A value of $x$ satisfying the equation $x^2 + b^2 = (a - x)^2$ is:

$\textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad  \textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad  \textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad  \textbf{(D)}\ \frac{a - b}{2}\qquad  \textbf{(E)}\ \frac{a^2 - b^2}{2}$

Solution

Solve for x:

\[x^2+b^2=(a-x)^2\] \[x^2+b^2=x^2-2ax+a^2\] \[b^2=-2ax+a^2\] \[2ax=a^2-b^2\] \[x=\frac{a^2-b^2}{2a} \to \boxed{\text{(C)}}\]


See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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