2005 AMC 12B Problem 12

Attempt Problem 12 of the 2005 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12B solutions, or check the answer key.

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12.

The quadratic equation x2+mx+n=0x^2 + mx + n = 0 has roots that are twice those of x2+px+m=0,x^2 + px + m = 0, and none of m,m, n,n, and pp is zero. What is the value of np?\dfrac{n}{p}?
11
22
44
88
1616
Answer: D
Concepts:Vieta’s Formulasquadratic
Small hint:
Let r1,r2r_1, r_2 be the roots of x2+px+m=0x^2 + px + m = 0; the other equation has roots 2r1,2r22r_1, 2r_2
Big hint:
Write m,n,pm, n, p in terms of r1r2r_1 r_2 and r1+r2r_1 + r_2 using Vieta’s formulas
Solution:
Let r1r_1 and r2r_2 be the roots of x2+px+m=0,x^2 + px + m = 0, so m=r1r2m = r_1 r_2 and p=−(r1+r2).p = -(r_1 + r_2).
The roots of x2+mx+n=0x^2 + mx + n = 0 are 2r12r_1 and 2r2,2r_2, so n=4r1r2n = 4r_1 r_2 and m=−2(r1+r2).m = -2(r_1 + r_2).
Then n=4mn = 4m and m=2p,m = 2p, which gives p=m2,p = \dfrac{m}{2}, so np=4mm2=8. \dfrac{n}{p} = \dfrac{4m}{\tfrac{m}{2}} = 8.
Thus, the correct answer is D.
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