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2008 AMC 10B Problem 9

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2008 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Quadratics · difficulty 1 of 5 · about 1 min · position P1-10

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

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

Answer

Key insight

By Vieta the roots sum to 2a/a = 2, so their average is 1 regardless of b.

Solution

For a quadratic $ax^2 + Bx + C = 0$ the roots sum to $-B/a$ . Here $B = -2a$ , so the two solutions add to
$-\frac{-2a}{a} = 2.$
Their average is $\tfrac{2}{2} = 1$ . The value of $b$ only affects whether the roots are real, not their average.

The answer is $\boxed{\textbf{(A)}\ 1}$ .

Why this works

The average of a quadratic's roots is the $x$ -coordinate of the vertex, $-B/(2a)$ , which depends only on the first two coefficients. Dividing by $a$ first, $x^2 - 2x + b/a = 0$ , makes the sum $2$ visible immediately. Vieta lets you answer "sum" or "average" questions without ever finding the roots.

Alternative approach

Complete the square: $a(x-1)^2 + (b - a) = 0$ , so the roots are $1 \pm \sqrt{(a-b)/a}$ , symmetric about $1$ .

The trap

Using the sign wrong in Vieta (sum = -2) or reporting the sum 2 instead of the average 1.

Common mistakes

Using the sign wrong in Vieta (sum = -2) or reporting the sum 2 instead of the average 1.
Trying to solve with the quadratic formula in terms of $a$ and $b$ and getting lost in the radical.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta