# 2008 AMC 10B Problems/Problem 9

## 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 1

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 $1\Rightarrow \boxed{A}$.

## Solution 2

We know that for an equation $ax^2 + bx + c = 0$, the sum of the roots is $\frac{-b}{a}$. This means that the sum of the roots for $ax^2 - 2ax + b = 0$ is $\frac{2a}{a}=2$. The average is the sum of the two roots divided by two, so the average is $\frac22 = 1 \Rightarrow \boxed{A}$.

## Solution 3

Using the quadratic formula(assuming $a=1$),we get $x = \frac{2 \pm \sqrt{4-4b}}{2}$. Let $y = \sqrt{4-4b}$. Simplifying, we get $1 \pm \frac{y}{2}$. That means the sum of the solutions is $(1+\frac{y}{2}) + (1-\frac{y}{2}) = 2$, so the average is $2/2 = \boxed{\textbf{(A) } 1}$.

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