AMC 10 · 2008 · #9

Grade 7 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

vieta-formulasquadratic-equationssymmetry-argument convert-to-algebra ↑ Prerequisites: quadratic-equations

📏 Medium solution 💡 1 insight

Problem

A quadratic equation ax² - 2ax + b = 0, whose leading coefficient a is not zero, is known to have two real solutions. What is the average of those two solutions?

Pick an answer.

(A)

$\ 1$

(B)

$\ 2$

(C)

$\ \frac ba$

(D)

$\ \frac{2b}a$

(E)

$\ \sqrt{2b-a}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

You are never told the two roots, and actually solving for them (with the quadratic formula and a messy √(·)) is far more work than the question needs. Tool #9 (Solve an Easier Related Problem) removes the clutter first: divide the whole equation by a so the leading coefficient becomes 1. Tool #4 (Introduce a Variable) is the key move — name the two unknown solutions r and s and write the equation as (x-r)(x-s)=0. Expanding that product shows that the coefficient of x is exactly -(r+s), which pins down the sum of the roots without ever finding them. Tool #3 (Eliminate Possibilities) then kills the choices that still depend on a or b, because the average turns out to be a plain constant.

1STEP 1

Divide by the leading coefficient

Since the equation is quadratic, a ≠ 0, so dividing every term by a leaves the roots untouched and makes the equation monic.

ax² - 2ax + b = 0 → x² - 2x + b/a = 0

2STEP 2

Name the roots and expand

Call the two roots r and s: the equation factors as (x-r)(x-s)=0, and expanding it makes the coefficient of x equal to -(r+s).

(x-r)(x-s) = x² - (r+s) x + rs

3STEP 3

Match coefficients to get the sum

Equal polynomials must match term by term, so -(r+s) = -2 and r+s = 2; the constant b/a never entered the sum.

-(r+s) = -2 → r+s = 2

4STEP 4

Average the roots and pick the answer

The average is half that sum: 1. No a or b appears, so choices built from them are out, and 2 is the sum, not the average.

average = (r+s)/2 = 2/2 = 1 → (A)

Answer

Test real quadratics of the required shape. With a=1, b=0: x² - 2x = 0 = x(x-2), roots 0 and 2, average (0+2)/2=1. With a=1, b=-3: x² - 2x - 3 = (x-3)(x+1)=0, roots 3 and -1, average (3+(-1))/2=1. Changing b moved the individual roots but the average stayed 1, exactly as the coefficient-matching argument predicted — which is why the a- and b-bearing choices (C), (D), (E) are traps and (A) is right.

💡Key takeaway

The two solutions of a quadratic balance around one point, and that balance point comes only from the first two coefficients — here it is 1, no matter what b is.

Divide by the leading coefficient
Name the roots and expand
Match coefficients to get the sum
Average the roots and pick the answer

Divide by the leading coefficient

Name the roots and expand

Match coefficients to get the sum

Average the roots and pick the answer

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