AMC 10 · 2008 · #9
Grade 7 algebraA quadratic equation ax2−2ax+b=0 has two real solutions. What is the average of these two solutions?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A quadratic equation $ax^2 - 2ax + b = 0$ is known to have two real solutions. Find the average of those two solutions, expressed using the coefficients $a$ and $b$.
Givens: The equation $ax^2 - 2ax + b = 0$ is quadratic, so its leading coefficient $a$ is not $0$; It has two real solutions, so the average of the two makes sense; The coefficient of $x$ is exactly $-2a$ and the constant term is $b$; Answer choices: (A) $1$, (B) $2$, (C) $\frac{b}{a}$, (D) $\frac{2b}{a}$, (E) $\sqrt{2b-a}$
Unknowns: The average $\frac{r+s}{2}$ of the two solutions $r$ and $s$
Understand
Restated: A quadratic equation $ax^2 - 2ax + b = 0$ is known to have two real solutions. Find the average of those two solutions, expressed using the coefficients $a$ and $b$.
Givens: The equation $ax^2 - 2ax + b = 0$ is quadratic, so its leading coefficient $a$ is not $0$; It has two real solutions, so the average of the two makes sense; The coefficient of $x$ is exactly $-2a$ and the constant term is $b$; Answer choices: (A) $1$, (B) $2$, (C) $\frac{b}{a}$, (D) $\frac{2b}{a}$, (E) $\sqrt{2b-a}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
You are never told the two roots, and actually solving for them (with the quadratic formula and a messy $\sqrt{\cdot}$) 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.
Execute — Answer: A
6.EE.A.3 Step 1 Divide by the leading coefficient
- Since the equation is quadratic, $a \neq 0$, so you may divide every term by $a$.
- Dividing an equation by a nonzero number never changes its solutions, so the roots stay exactly the same.
- The equation becomes monic (leading coefficient $1$), which makes its coefficients easy to read.
💡 Scaling every term of an equation by the same nonzero number leaves its solutions untouched.
6.EE.A.3 Step 2 Name the roots and expand
- Call the two solutions $r$ and $s$.
- A monic quadratic with those roots can be written in factored form $(x-r)(x-s)=0$.
- Multiply it out with the distributive property to compare it to $x^2 - 2x + \frac{b}{a}$.
- The expansion shows the coefficient of $x$ is $-(r+s)$ and the constant is $rs$.
💡 Expanding $(x-r)(x-s)$ turns the hidden roots into visible coefficients you can match.
7.NS.A.1 Step 3 Match coefficients to get the sum
- The two forms $x^2 - (r+s)x + rs$ and $x^2 - 2x + \frac{b}{a}$ are the same polynomial, so their matching parts must be equal.
- Comparing the coefficient of $x$ gives $-(r+s) = -2$, hence $r+s = 2$.
- Notice this used only the middle coefficient — the constant $\frac{b}{a}$ never entered the sum.
💡 If two equal polynomials are lined up, the number sitting in front of $x$ must agree on both sides.
6.EE.A.2 Step 4 Average the roots and pick the answer
- The average of the two solutions is their sum divided by $2$: $\frac{r+s}{2} = \frac{2}{2} = 1$.
- This value is a plain constant — it does not contain $a$ or $b$ at all.
- So every choice that still shows $a$ or $b$, namely (C) $\frac{b}{a}$, (D) $\frac{2b}{a}$, and (E) $\sqrt{2b-a}$, cannot be right, and (B) $2$ is the sum, not the average.
- The average is $1$, which is choice (A).
💡 The midpoint of the two roots is half their sum, and that sum came out to a fixed number.
6.EE.A.3 Since the equation is quadratic, $a \neq 0$, so you may divide every term by $a$ 6.EE.A.3 Call the two solutions $r$ and $s$. A monic quadratic with those roots can be wr 7.NS.A.1 The two forms $x^2 - (r+s)x + rs$ and $x^2 - 2x + \frac{b}{a}$ are the same poly 6.EE.A.2 The average of the two solutions is their sum divided by $2$: $\frac{r+s}{2} = \ Review
Reasonableness: Test real quadratics of the required shape. With $a=1,\,b=0$: $x^2 - 2x = 0 = x(x-2)$, roots $0$ and $2$, average $\frac{0+2}{2}=1$. With $a=1,\,b=-3$: $x^2 - 2x - 3 = (x-3)(x+1)=0$, roots $3$ and $-1$, average $\frac{3+(-1)}{2}=1$. 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.
Alternative: Think of the graph $y = ax^2 - 2ax + b$, a parabola that crosses the $x$-axis at the two roots. A parabola is symmetric about a vertical line, so the two crossing points are mirror images and their average is the $x$-coordinate of that line of symmetry, $x = \frac{-(-2a)}{2a} = 1$. Drawing the picture (Tool #1) gives the average $1$ directly, with no algebra.
CCSS standards used (min grade 7)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Dividing the equation by $a$ to make it monic, and expanding $(x-r)(x-s)$ with the distributive property to compare coefficients.)7.NS.A.1Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Matching the middle coefficients $-(r+s) = -2$ and handling the signs to conclude $r+s = 2$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading the coefficients $-2a$ and $b$, and evaluating the average $\frac{r+s}{2}$ from the known sum of the roots.)
⭐ 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.
⭐ 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.
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