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Difference between revisions of "1962 AHSME Problems/Problem 19"

(→‎Solution)
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==Solution==
 
==Solution==
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{{solution}}
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Substituting in the <math>(x, y)</math> pairs gives the following system of equations:
 +
<cmath>a-b+c=12</cmath>
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<cmath>c=5</cmath>
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<cmath>4a+2b+c=-3</cmath>
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We know <math>c=5</math>, so plugging this in reduces the system to two variables:
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<cmath>a-b=7</cmath>
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<cmath>4a+2b=-8</cmath>
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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 $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:

$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution

Substituting in the $(x, y)$ pairs gives the following system of equations: \[a-b+c=12\] \[c=5\] \[4a+2b+c=-3\] We know $c=5$, so plugging this in reduces the system to two variables: \[a-b=7\] \[4a+2b=-8\] Dividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation to get $3a=3$, or $a=1$. So the solution set is $(1, -6, 5)$, and the sum is $\boxed{0\textbf{ (C)}}$.

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