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

Latest revision as of 18:06, 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 $\boxed{\textbf{(A)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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
All AHSME Problems and Solutions

@@ Line 22: / Line 22: @@
 ==See Also==
-{{AHSME box|year=1960|num-b=21|num-a=23}}
+{{AHSME 40p box|year=1960|num-b=21|num-a=23}}
+[[Category:Introductory Algebra Problems]]

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

Latest revision as of 18:06, 17 May 2018

Problem 22

Solution

See Also