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2005 AMC 12B Problem 12

Problem 12 of 25IntermediateAlgebra

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}?

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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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Concepts: Vieta’s Formulas · quadratic

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