Difference between revisions of "1960 AHSME Problems/Problem 9"

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==See Also==
 
{{AHSME 40p box|year=1960|num-b=8|num-a=10}}
 
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:58, 17 May 2018

Problem

The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c):

$\text{(A) irreducible}\qquad$

$\text{(B) reducible to negative 1}\qquad$

$\text{(C) reducible to a polynomial of three terms}\qquad$

$\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$

$\text{(E) reducible to} \frac{a+b-c}{a-b+c}$

Solution

Use the commutative property to get \[\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\] Factor perfect square trinomials to get \[\frac{(a+b)^2-c^2}{(a+c)^2-b^2}\] Factor difference of squares to get \[\frac{(a+b+c)(a+b-c)}{(a+b+c)(a-b+c)}\] Cancel out like terms (with suitable restrictions of a, b, c) to get \[\frac{a+b-c}{a-b+c}\] The answer is $\boxed{\textbf{(E)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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