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

Revision as of 15:35, 16 January 2021

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 $n/p$ ?

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

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

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 $2a$ 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 $\frac{n}{p} = 8 = \boxed{\textbf{D}}$

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 5: / Line 5: @@
 <math>\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(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 18: / Line 19: @@
 Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
-== Solution 2 ==
+===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
+If the roots of <math>x^2 + mx + n = 0</math> are <math>2a</math> and <math>2b</math> and the roots of <math>x^2 + px + m = 0</math> are <math>a</math> and <math>b</math>, then using Vieta's formulas,
+<cmath>2a + 2b = -m</cmath>
+<cmath>a + b = -p</cmath>
+<cmath>2a(2b) = n</cmath>
+<cmath>a(b) = m</cmath>
+Therefore, substituting the second equation into the first equation gives
+<cmath>m = 2(p)</cmath>
+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>
-Using Vieta's formula, the sum of the roots of <math>x^2 + mx + n</math> is <math>-m</math> and the product of the roots is <math>n</math>
+== Video Solution ==
+https://youtu.be/3dfbWzOfJAI?t=1023
-The sum of the roots of <math>x^2 + px + m</math> is <math>-p</math> and the product of the roots is <math>m</math>
+~ pi_is_3.14
-We now have two equations:
-<math>-2p = -m</math>
-and <math>4m = n</math>
-Solving the first equation for p, we have
-<math>p = \frac{m}{2}</math>
-Thus $\frac{n}{p} = \frac{4m}{\frac{m}{2}} = 8
 == See also ==
@@ Line 42: / Line 41: @@
 * [[Vieta's Formulas]]
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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