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

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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:
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<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>.
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 18|num-a=20}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:17, 3 October 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 $\boxed{0\textbf{ (C)}}$.

See Also

1962 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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