Difference between revisions of "1960 AHSME Problems/Problem 22"

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{{AHSME 40p box|year=1960|num-b=21|num-a=23}}
 
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:06, 17 May 2018

Problem 22

The equality $(x+m)^2-(x+n)^2=(m-n)^2$, where $m$ and $n$ are unequal non-zero constants, is satisfied by $x=am+bn$, where:

$\textbf{(A)}\ a = 0, b \text{ } \text{has a unique non-zero value}\qquad \\ \textbf{(B)}\ a = 0, b \text{ } \text{has two non-zero values}\qquad \\ \textbf{(C)}\ b = 0, a \text{ } \text{has a unique non-zero value}\qquad \\ \textbf{(D)}\ b = 0, a \text{ } \text{has two non-zero values}\qquad \\ \textbf{(E)}\ a \text{ } \text{and} \text{ } b \text{ } \text{each have a unique non-zero value}$

Solution

Expand binomials, combine like terms, and subtract terms from both sides. \[x^2 + 2xm + m^2 - x^2 - 2xn - n^2 = m^2 - 2mn + n^2\] \[2xm + m^2 - 2xn - n^2 = m^2 - 2mn + n^2\] \[2xm - 2xn - n^2 = -2mn + n^2\] Get all the x-terms on one side and factor to solve for x. \[2xm - 2xn = -2mn + 2n^2\] \[2x(m-n) = -2n(m-n)\] Since $m \not= n$, both sides can be divided by $m-n$. \[x = -n\] That means $a = 0$ and $b = -1$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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