AMC 2005 12B P12

Problem Statement

The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$, and none of $m,n,$ and $p$ is zero. What is the value of $n/p$?

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Solution

$$f(x) = x^2+mx+n = (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1r_2 \implies - (r_1 + r_2) = m, r_1r_2 = n$$ $$g(x) = x^2+px+m = (x - r_3)(x - r_4) = x^2 - (r_3 + r_4)x + r_3r_4 \implies - (r_3 + r_4) = p, r_3r_4 = m$$ $$r_1 = 2r_3, r_2 = 2r_4$$ $$r_1r_2 = (2r_3)(2r_4) = 4r_3r_4 = 4m$$ $$-(r_3 + r_4) = \dfrac {1}{2}(m)$$ $$\dfrac {n}{p} = \dfrac {r_1r_2}{-(r_3 + r_4)}$$ $$\therefore \dfrac {n}{p} = \dfrac {4m}{\dfrac {1}{2}(m)} = 8$$