Difference between revisions of "2005 AMC 12B Problems/Problem 12"
Pi is 3.14 (talk | contribs) (→Solution) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
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<math>\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}</math> | <math>\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}</math> | ||
− | == Video Solution == | + | === Video Solution === |
https://youtu.be/3dfbWzOfJAI?t=1023 | https://youtu.be/3dfbWzOfJAI?t=1023 | ||
Revision as of 03:25, 14 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 has roots twice those of
, and none of
and
is zero. What is the value of
?
Video Solution
https://youtu.be/3dfbWzOfJAI?t=1023
~ pi_is_3.14
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
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.