Difference between revisions of "1950 AHSME Problems/Problem 4"
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We can further factor to get <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> If we assume b is not equal to a, this is equal to <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b}{a}=\dfrac{(a-b)(a+b)}{ab}-\dfrac{b^2}{ab}=\dfrac{(a-b)(a+b)-b^2}{ab}=\dfrac{a^2-2b^2}{ab}</math> | We can further factor to get <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> If we assume b is not equal to a, this is equal to <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b}{a}=\dfrac{(a-b)(a+b)}{ab}-\dfrac{b^2}{ab}=\dfrac{(a-b)(a+b)-b^2}{ab}=\dfrac{a^2-2b^2}{ab}</math> | ||
| − | The answer is | + | The answer is \boxed{\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}} |
==See Also== | ==See Also== | ||
{{AHSME box|year=1950|num-b=3|num-a=5}} | {{AHSME box|year=1950|num-b=3|num-a=5}} | ||
Revision as of 13:48, 28 October 2011
Problem
Reduced to lowest terms, 
- 
is equal to:
Solution
We can use difference of two squares to expand 
and factor to get 
and 
We can further factor to get 
If we assume b is not equal to a, this is equal to 
The answer is \boxed{\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}}
See Also
| 1950 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
