Difference between revisions of "2005 AMC 12B Problems/Problem 12"
(→Problem) |
|||
(25 intermediate revisions by 13 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #12]] and [[2005 AMC 10B Problems|2005 AMC 10B #16]]}} | ||
== Problem == | == Problem == | ||
+ | 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>? | ||
− | == Solution == | + | <math>\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math> |
+ | |||
+ | ==Solutions== | ||
+ | ===Solution 1=== | ||
+ | Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then | ||
+ | |||
+ | <cmath>x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,</cmath> | ||
+ | |||
+ | so <math>p = -(a+b)</math> and <math>m = ab</math>. Also, <math>x^2 + mx + n = 0</math> has roots <math>2a</math> and <math>2b</math>, so | ||
+ | |||
+ | <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{\textbf{(D) }8}</math>. | ||
+ | |||
+ | Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | 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}= \boxed{\textbf{(D) }8}</math> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/3dfbWzOfJAI?t=1023 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
− | * [[ | + | {{AMC10 box|year=2005|ab=B|num-b=15|num-a=17}} |
+ | {{AMC12 box|year=2005|num-b=11|num-a=13|ab=B}} | ||
+ | * [[Vieta's Formulas]] | ||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:30, 24 August 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 has roots twice those of
, and none of
and
is zero. What is the value of
?
Solutions
Solution 1
Let have roots
and
. Then
so and
. Also,
has roots
and
, so
and and
. Thus
.
Indeed, consider the quadratics .
Solution 2
If the roots of are
and
and the roots of
are
and
, then using Vieta's formulas,
Therefore, substituting the second equation into the first equation gives
and substituting the fourth equation into the third equation gives
Therefore,
, so
Video Solution
https://youtu.be/3dfbWzOfJAI?t=1023
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.