Difference between revisions of "1988 AHSME Problems/Problem 19"
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==Solution 4 (fastest)== | ==Solution 4 (fastest)== | ||
After regrouping, the numerator becomes <math>(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2</math>. Factoring further, we get <math>(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)</math>. After dividing, we get <math>a^2x^2+b^2y^2+2bxay</math>, which can be factored as <math>(ax+by)^2</math>, so the answer is <math>\boxed{\text{B}}</math>. | After regrouping, the numerator becomes <math>(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2</math>. Factoring further, we get <math>(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)</math>. After dividing, we get <math>a^2x^2+b^2y^2+2bxay</math>, which can be factored as <math>(ax+by)^2</math>, so the answer is <math>\boxed{\text{B}}</math>. | ||
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-Pengu14 | -Pengu14 | ||
Latest revision as of 20:47, 16 January 2024
Problem
Simplify
Solution 1
We can multiply each answer choice by 
and then compare with the numerator. This gives 
.
Solution 2
Expanding everything in the brackets, we get
. We can then group numbers up in pairs so they equal 
:
We get .
-ThisUsernameIsTaken
Solution 3
If you were out of time and your algebra isn't that good, you could just plug in some values for the variables and see which answer choice works.
Solution 4 (fastest)
After regrouping, the numerator becomes 
. Factoring further, we get 
. After dividing, we get 
, which can be factored as 
, so the answer is .
-Pengu14
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
