AMC 10 · 2006 · #14
Grade 7 algebraLet a and b be the roots of the equation x2−mx+2=0. Suppose that a+b1 and b+a1 are the roots of the equation x2−px+q=0. What is q?
Pick an answer.
AMC 10 2006 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: Two numbers $a$ and $b$ are the solutions of $x^2-mx+2=0$. Build two new numbers $a+\frac1b$ and $b+\frac1a$; these are the solutions of $x^2-px+q=0$. Find $q$, which is the value that plays the role of the constant term in that second equation.
Givens: $a$ and $b$ are the roots of $x^2-mx+2=0$; $a+\frac1b$ and $b+\frac1a$ are the roots of $x^2-px+q=0$; Answer choices: (A) $\frac52$, (B) $\frac72$, (C) $4$, (D) $\frac92$, (E) $8$
Unknowns: The value of $q$
Understand
Restated: Two numbers $a$ and $b$ are the solutions of $x^2-mx+2=0$. Build two new numbers $a+\frac1b$ and $b+\frac1a$; these are the solutions of $x^2-px+q=0$. Find $q$, which is the value that plays the role of the constant term in that second equation.
Givens: $a$ and $b$ are the roots of $x^2-mx+2=0$; $a+\frac1b$ and $b+\frac1a$ are the roots of $x^2-px+q=0$; Answer choices: (A) $\frac52$, (B) $\frac72$, (C) $4$, (D) $\frac92$, (E) $8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
The equation $x^2-px+q=0$ hides $q$ as the product of its two roots, so Tool #4 (Introduce a Variable) is used the smart way: instead of hunting for the messy roots $a$ and $b$ themselves, treat their product $ab$ as a single quantity that Vieta's relation hands us for free from the first equation. Tool #16 (Change Focus) is the key move — we shift attention away from "what are $a$ and $b$?" toward "what is $q$ as a product?", and it turns out the product collapses to something built only from $ab$. Tool #7 (Identify Subproblems) splits the work into two clean pieces: first read $ab$ off the first equation, then expand the product that defines $q$.
Execute — Answer: D
6.EE.A.3 Step 1 Read the product ab off the first equation
- A monic quadratic with roots $a$ and $b$ factors as $(x-a)(x-b)=x^2-(a+b)x+ab$.
- Matching this against $x^2-mx+2$ term by term, the constant terms must agree, so $ab=2$.
- (The middle terms give $a+b=m$, but we will not even need that.)
💡 Two quadratics that share the same roots must match coefficient by coefficient, so the constant term just is the product of the roots.
6.EE.A.2 Step 2 See q as a product, not as two roots
- In $x^2-px+q=0$ the same rule applies: $q$ is the product of that equation's two roots.
- Those roots are $a+\frac1b$ and $b+\frac1a$, so $q=\left(a+\frac1b\right)\left(b+\frac1a\right)$.
- This is the whole point — we never solve for $a$ and $b$; we only need to multiply these two expressions together.
💡 The constant term of a monic quadratic is always the product of its roots, so finding $q$ means multiplying, not solving.
6.EE.A.3 Step 3 Expand the product
- Multiply the two brackets using the distributive property, one pair of terms at a time: $a\cdot b$, then $a\cdot\frac1a=1$, then $\frac1b\cdot b=1$, then $\frac1b\cdot\frac1a=\frac1{ab}$.
- The two middle pieces each simplify to $1$, so the whole thing depends only on $ab$ and its reciprocal.
💡 The cross terms pair a quantity with its own reciprocal, so they collapse to plain $1$s and only the product $ab$ survives.
7.NS.A.3 Step 4 Substitute ab = 2
- From Step 1, $ab=2$, so its reciprocal is $\frac12$.
- Substitute both into the expanded form and add the rational numbers: $2+\frac12+2=\frac92$.
- So $q=\frac92$, which is choice (D).
💡 Once every letter is replaced by the known product, the answer is just a short fraction sum.
6.EE.A.3 A monic quadratic with roots $a$ and $b$ factors as $(x-a)(x-b)=x^2-(a+b)x+ab$. 6.EE.A.2 In $x^2-px+q=0$ the same rule applies: $q$ is the product of that equation's two 6.EE.A.3 Multiply the two brackets using the distributive property, one pair of terms at 7.NS.A.3 From Step 1, $ab=2$, so its reciprocal is $\frac12$. Substitute both into the ex Review
Reasonableness: The answer never used $m$, which is exactly right: the problem left $m$ free, so a correct value of $q$ had to be independent of it — and $\frac92$ is. A concrete sanity check: pick $ab=2$ with $a=b=\sqrt2$ (the case $m=2\sqrt2$). Then $a+\frac1b=\sqrt2+\frac1{\sqrt2}=\frac{3}{\sqrt2}$ and likewise for the other root, so their product is $\frac{3}{\sqrt2}\cdot\frac{3}{\sqrt2}=\frac92$, matching. The value sits sensibly between choices (C) $4$ and (E) $8$, and equals (D).
Alternative: Instead of using Vieta on the second equation, you could compute $q$ as the product of the roots directly after rewriting each root over a common denominator: $a+\frac1b=\frac{ab+1}{b}$ and $b+\frac1a=\frac{ab+1}{a}$, so $q=\frac{(ab+1)^2}{ab}$. With $ab=2$ this is $\frac{(2+1)^2}{2}=\frac{9}{2}$ — the same answer, and it shows cleanly why only $ab$ matters.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Treating $a$, $b$, $m$, $q$, and the product $ab$ as symbols, and reading $q$ as the product of the second equation's roots.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x-a)(x-b)$ to match coefficients (getting $ab=2$) and expanding $\left(a+\frac1b\right)\left(b+\frac1a\right)$ with the distributive property.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Substituting $ab=2$ and adding the rational numbers $2+\frac12+2$ to get $\frac92$.)
⭐ The constant term of $x^2-px+q$ is just the product of its roots, so multiply $a+\frac1b$ by $b+\frac1a$ — it collapses to $ab+\frac1{ab}+2$, and $ab=2$ gives $\frac92$.
⭐ The constant term of $x^2-px+q$ is just the product of its roots, so multiply $a+\frac1b$ by $b+\frac1a$ — it collapses to $ab+\frac1{ab}+2$, and $ab=2$ gives $\frac92$.
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