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Difference between revisions of "2008 AMC 10B Problems/Problem 9"

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Revision as of 11:26, 4 July 2013

Problem

A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of these two solutions?

$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}$

Solution

Dividing both sides by $a$, we get $x^2 - 2x + b/a = 0$. By Vieta's formulas, the sum of the roots is $2$, therefore their average is $1\Rightarrow \boxed{A}$.

Another Solution

We know that for an equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$. This means that the sum of the roots for $ax^2 - 2ax + b = 0$ is $2a/a$, or 2. The average is the sum of the two roots divided by two, so the average is $2/2 = 1$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 10 Problems and Solutions

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