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2005 AMC 10B Problem 16

2005 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Quadratics · difficulty 3 of 5 · about 3 min · position P16-20

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 $n/p$ ?

A)
${{{1}}}$
B)
${{{2}}}$
C)
${{{4}}}$
D)
${{{8}}}$
E)
${{{16}}}$

Answer

Key insight

Doubling both roots doubles the sum and quadruples the product, so Vieta gives m = 2p and n = 4m, hence n = 8p.

Solution

Call the roots of $x^2 + px + m$ by $r$ and $s$ . Vieta's formulas give
$r + s = -p, \qquad rs = m.$

The other quadratic, $x^2 + mx + n$ , has roots $2r$ and $2s$ , so
$2r + 2s = -m, \qquad (2r)(2s) = n.$

Compare: the sum equation says $-m = 2(r+s) = -2p$ , so $m = 2p$ . The product equation says $n = 4rs = 4m$ .

Chain them: $n = 4m = 4(2p) = 8p$ , and since $p \ne 0$ ,
$\frac{n}{p} = 8.$

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

A statement about roots is a statement about sums and products via Vieta. Scaling every root by $k$ multiplies the sum by $k$ and the product by $k^2$ , which is exactly why the two coefficients pick up factors of $2$ and $4$ . The shared coefficient $m$ is the bridge that links the two equations.

Alternative approach

Sanity check with a concrete pair: $x^2 + 8x + 16$ has double root $-4$ ; doubling gives root $-8$ , i.e. $x^2 + 16x + 64$ . Here $p = 8, m = 16, n = 64$ and $n/p = 8$ .

The trap

Mixing up which coefficient is a sum and which is a product, e.g. concluding n = 2m or m = 4p and answering 4 or 16.

Common mistakes

Mixing up which coefficient is a sum and which is a product, e.g. concluding n = 2m or m = 4p and answering 4 or 16.
Dropping the sign in $r + s = -p$ on one equation but not the other, producing $m = -2p$ .

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Substitute to simplify (u = x+1/x, shifting, scaling)