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2005 AMC 12B Problems/Problem 12

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Problem

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?

\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(...

Solution

Let x^2 + px + m = 0 have roots a and b. Then

x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,

so p = -(a+b) and m = ab. Also, x^2 + mx + n = 0 has roots 2a and 2b, so

x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,

and m = -2(a+b) and n = 4ab. Thus \frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = 8 \Longrightarrow \textbf{(D)}.

Indeed, consider the quadratics x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0. See Vieta's formulas.

See also

2005 AMC 12B (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
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