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Difference between revisions of "2005 AMC 12B Problems/Problem 12"
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We now have two equations:
 
We now have two equations:
  
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<math>-2p = -m</math>
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<cmath>-2p = -m,</cmath>
  
 
and <math>4m = n</math>
 
and <math>4m = n</math>

Revision as of 00:31, 31 October 2012

The following problem is from both the 2005 AMC 12B #12 and 2005 AMC 10B #16, so both problems redirect to this page.

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 $n/p$?

$\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}$

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 $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\mathrm{(D)}\ 8}$.

Indeed, consider the quadratics $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$.

Solution 2

Realize that if first quadratic has roots twice those of the second, then the sum of its roots will be twice those of the second and the product of its roots will be four times those of the second

Using Vieta's formula, the sum of the roots of $x^2 + mx + n$ is $-m$ and the product of the roots is $n$

The sum of the roots of $x^2 + px + m$ is $-p$ and the product of the roots is $m$

We now have two equations:

\[-2p = -m,\]

and $4m = n$

Solving the first equation for p, we have

$p = \frac{m}{2}$

Thus $\frac{n}{p} = \frac{4m}{\frac{m}{2}} = 8

See also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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