AMC 10 · 2006 · #14

Grade 7 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta
vieta-formulasquadratic-equationspolynomial-roots convert-to-algebra ↑ Prerequisites: quadratic-equationsvieta-formulas
📏 Medium solution 💡 3 insights
Problem
Two numbers a and b are the solutions of x²-mx+2=0, where m is some unspecified constant. From them build two new numbers, a+1b\frac{1}{b} and b+1a\frac{1}{a}, and those two are the solutions of x²-px+q=0. Find q, the value that plays the role of the constant term in that second equation.

Pick an answer.

(A)
$\frac{5}{2}$
(B)
$\frac{7}{2}$
(C)
4
(D)
$\frac{9}{2}$
(E)
8

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The equation x²-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.

1STEP 1

Read the product ab off the first equation

Match (x-a)(x-b)=x²-(a+b)x+ab against x²-mx+2 term by term; the constant terms give ab=2.

(x-a)(x-b)=x²-(a+b)x+ab=x²-mx+2 → ab=2
2STEP 2

See q as a product, not as two roots

The same rule reads q off x²-px+q=0: q=(a+1b\frac{1}{b})(b+1a\frac{1}{a}), a product to multiply, not roots to solve for.

q=(a+dfrac1b)(b+dfrac1a)
3STEP 3

Expand the product

Distribute: ab, then a·1a\frac{1}{a}=1, then 1b\frac{1}{b}·b=1, then 1ab\frac{1}{ab} — so q=ab+1ab\frac{1}{ab}+2.

q=ab+a·dfrac1a+dfrac1b · b+1/ab=ab+1+1+1/ab=ab+1/ab+2
4STEP 4

Substitute ab = 2

With ab=2 the reciprocal is 12\frac{1}{2}, so q=2+12\frac{1}{2}+2=92\frac{9}{2}, choice (D).

q=ab+1/ab+2=2+1/2+2=9/2 → (D)
Answer
9/2
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 92\frac{9}{2} is. A concrete sanity check: pick ab=2 with a=b=2\sqrt{2} (the case m=22\sqrt{2}). Then a+1b\frac{1}{b}=2+12=32\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}} and likewise for the other root, so their product is 32⋅32=92\frac{3}{\sqrt{2}}\cdot\frac{3}{\sqrt{2}}=\frac{9}{2}, matching. The value sits sensibly between choices (C) 4 and (E) 8, and equals (D).
💡Key takeaway

The constant term of x²-px+q is just the product of its roots, so multiply a+1b\frac{1}{b} by b+1a\frac{1}{a} — it collapses to ab+1ab\frac{1}{ab}+2, and ab=2 gives 92\frac{9}{2}.

  • Read the product ab off the first equation
  • See q as a product, not as two roots
  • Expand the product
  • Substitute ab = 2

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