1962 AHSME Problems/Problem 19: Difference between revisions

Revision as of 10:37, 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 $\boxed{0\textbf{ (C)}}$ .

@@ Line 5: / Line 5: @@
 ==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>.

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1962 AHSME Problems/Problem 19: Difference between revisions

Revision as of 10:37, 17 April 2014

Problem

Solution