Difference between revisions of "2005 AMC 12B Problems/Problem 12"

Latest revision as of 01:57, 11 November 2024

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 $n/p$ ?

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

Solutions

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

To test that this actually works, 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 $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, $2a + 2b = -m$ $a + b = -p$ $2a(2b) = n$ $a(b) = m$ Therefore, substituting the second equation into the first equation gives $m = 2(p)$ and substituting the fourth equation into the third equation gives $n = 4(m)$ Therefore, $n = 8p$ , so $\frac{n}{p}= \boxed{\textbf{(D) }8}$

Video Solution

https://youtu.be/3dfbWzOfJAI?t=1023

~ pi_is_3.14

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

Vieta's Formulas

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 3: / Line 3: @@
 The [[quadratic equation]] <math>x^2+mx+n</math> has roots twice those of <math>x^2+px+m</math>, and none of <math>m,n,</math> and <math>p</math> is zero. What is the value of <math>n/p</math>?
-<math>\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}</math>
+<math>\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math>
-== Solution ==
+==Solutions==
 ===Solution 1===
 Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then
@@ Line 17: / Line 15: @@
 <cmath>x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,</cmath>
-and <math>m = -2(a+b)</math> and <math>n = 4ab</math>. Thus <math>\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\mathrm{(D)}\ 8}</math>.
+and <math>m = -2(a+b)</math> and <math>n = 4ab</math>. Thus <math>\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\textbf{(D) }8}</math>.
-Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
+To test that this actually works, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
 ===Solution 2===
@@ Line 31: / Line 29: @@
 and substituting the fourth equation into the third equation gives
 <cmath>n = 4(m)</cmath>
-Therefore, <math>n = 8p</math>, so <math>\frac{n}{p} = 8 = \boxed{\textbf{D}}</math>
+Therefore, <math>n = 8p</math>, so <math>\frac{n}{p}= \boxed{\textbf{(D) }8}</math>
+== Video Solution ==
+https://youtu.be/3dfbWzOfJAI?t=1023
+~ pi_is_3.14
 == See also ==

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