AMC 10 · 2005 · #16

Grade 8 algebra
Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Chained Ratio Linking
quadratic-equationsvieta-formulaspolynomial-roots convert-to-algebravieta-formulas ↑ Prerequisites: quadratic-equationsvieta-formulaspolynomial-roots
📏 Medium solution 💡 2 insights
Problem
The quadratic x²+mx+n has roots that are exactly twice the roots of x²+px+m. None of m, n, p is zero. Find the value of n/p.

Pick an answer.

(A)
1
(B)
2
(C)
4
(D)
8
(E)
16

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The roots are the hidden link between the two quadratics, so Tool #4 (Introduce a Variable) names them r and s. A quadratic built from roots r,s factors as (x-r)(x-s)=x²-(r+s)x+rs, so each coefficient is just the sum or the product of the roots. Tool #13 (Convert to Algebra) turns 'twice the roots' into equations by matching coefficients. Tool #7 (Identify Subproblems) splits the work into two clean channels: what the sums of roots tell us, and separately what the products tell us. Each channel gives one link (m=2p and n=4m), and chaining them answers n/p.

1STEP 1

Name the roots of the first quadratic

Let the roots of x²+px+m be r and s. Matching (x-r)(x-s)=x²-(r+s)x+rs term by term gives r+s=-p and rs=m.

(x-r)(x-s)=x²-(r+s)x+rs → r+s=-p, rs=m
2STEP 2

Name the roots of the second quadratic

The second quadratic's roots are 2r and 2s, so (x-2r)(x-2s)=x²-2(r+s)x+4rs gives 2r+2s=-m and 4rs=n.

2r+2s=-m, (2r)(2s)=4rs=n
3STEP 3

Bridge the two through the shared roots

Substitute r+s=-p into 2r+2s=-m to get m=2p, and rs=m into n=4rs to get n=4m.

-m=2(r+s)=-2p → m=2p; n=4rs=4m
4STEP 4

Chain the links to get n/p

Chaining n=4m with m=2p gives n=4(2p)=8p, and dividing by the nonzero p leaves n/p=8 — choice (D).

n=4m=4(2p)=8p → n/p=8 → (D)
Answer
8
The individual roots r,s never appear in the final answer, so the ratio n/p=8 should hold for any valid pair of quadratics — a good sign the algebra is clean. Build a concrete example to confirm: take p=8. Then m=2p=16 and n=8p=64. The first quadratic is x²+8x+16=(x+4)², with a double root at -4. Its roots doubled are -8,-8, and (x+8)²=x²+16x+64 does equal x²+mx+n with m=16, n=64. Indeed n/p=64/8=8, matching (D). The trap choice (E) 16 comes from doubling twice (once for the sum, once again for the product) instead of using the product's factor of 4; choice (C) 4 comes from forgetting the extra factor of 2 hidden in m=2p.
💡Key takeaway

A quadratic's coefficients are just the sum and product of its roots in disguise, so naming the roots turns 'twice the roots' into simple equations you can chain together.

  • Name the roots of the first quadratic
  • Name the roots of the second quadratic
  • Bridge the two through the shared roots
  • Chain the links to get n/p

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