Difference between revisions of "1962 AHSME Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | {{ | + | Substituting in the <math>(x, y)</math> pairs gives the following system of equations: |
+ | <cmath>a-b+c=12</cmath> | ||
+ | <cmath>c=5</cmath> | ||
+ | <cmath>4a+2b+c=-3</cmath> | ||
+ | We know <math>c=5</math>, so plugging this in reduces the system to two variables: | ||
+ | <cmath>a-b=7</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>. |
Revision as of 11:37, 17 April 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
.