Difference between revisions of "1962 AHSME Problems/Problem 19"
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<cmath>4a+2b=-8</cmath> | <cmath>4a+2b=-8</cmath> | ||
Dividing the second equation by 2 gives <math>2a+b=-4</math>, which can be added to the first equation to get <math>3a=3</math>, or <math>a=1</math>. So the solution set is <math>(1, -6, 5)</math>, and the sum is <math>\boxed{0\textbf{ (C)}}</math>. | Dividing the second equation by 2 gives <math>2a+b=-4</math>, which can be added to the first equation to get <math>3a=3</math>, or <math>a=1</math>. So the solution set is <math>(1, -6, 5)</math>, and the sum is <math>\boxed{0\textbf{ (C)}}</math>. | ||
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+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1962|before=Problem 18|num-a=20}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:17, 3 October 2014
Problem
If the parabola passes through the points , , and , the value of is:
Solution
Substituting in the pairs gives the following system of equations: We know , so plugging this in reduces the system to two variables: Dividing the second equation by 2 gives , which can be added to the first equation to get , or . So the solution set is , and the sum is .
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |
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