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Difference between revisions of "2005 AMC 12B Problems/Problem 12"
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== Solution ==
 
== Solution ==
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===Solution 1===
 
Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then  
 
Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then  
  
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Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
 
Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
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===Solution 2===
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If the roots of <math>x^2 + mx + n = 0</math> are twice the roots of <math>x^2 + px + m = 0</math>, then the sum of the roots of the first equation is also twice the sum of the roots of the second equation, and the product of the roots of the first equation is <math>2(2) = 4</math> times the product of the roots of the second equation. Using Vieta's equations, <math>-m = 2(-p)</math>, and <math>n = 4(m)</math>. Substituting, <math>n = 8p</math>,  so <math>\frac{n}{p} = 8 = \boxed{\textbf{D}}</math>
  
 
== See also ==
 
== See also ==

Revision as of 11:17, 22 February 2019

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

Solution 1

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

If the roots of $x^2 + mx + n = 0$ are twice the roots of $x^2 + px + m = 0$, then the sum of the roots of the first equation is also twice the sum of the roots of the second equation, and the product of the roots of the first equation is $2(2) = 4$ times the product of the roots of the second equation. Using Vieta's equations, $-m = 2(-p)$, and $n = 4(m)$. Substituting, $n = 8p$, so $\frac{n}{p} = 8 = \boxed{\textbf{D}}$

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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