Difference between revisions of "2017 AMC 12A Problems/Problem 23"
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<math>f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{\textbf{(C)}\,-7007}</math> | <math>f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{\textbf{(C)}\,-7007}</math> | ||
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+ | ==Solution 3== | ||
+ | Let the roots of <math>g(x)</math> be <math>p,q,r</math> and the roots of <math>f(x)</math> be <math>p,q,r,s</math>. Then by Vietas, <cmath>-100=pqr+pqs+prs+qrs = -10+ s(pq+pr+rs) = -10 + s,</cmath>so <math>s = -90</math>. Again by Vietas, <math>p+q+r+s = -a + s = -1 \implies a = -89</math>. Finally, <math>f(1) = (1-(-90))g(1) = \textbf{(C)}\,-7007</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 17:02, 23 May 2020
Problem
For certain real numbers , , and , the polynomial has three distinct roots, and each root of is also a root of the polynomial What is ?
Solution
Let and be the roots of . Let be the additional root of . Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:
Thus .
Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:
Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:
It follows that . But so
Now we can factor in terms of as
Then and
Hence .
Solution 2
Since all of the roots of are distinct and are roots of , and the degree of is one more than the degree of , we have that
for some number . By comparing coefficients, we see that . Thus,
Expanding and equating coefficients we get that
The third equation yields , and the first equation yields . So we have that
Solution 3
Let the roots of be and the roots of be . Then by Vietas, so . Again by Vietas, . Finally, .
Video Solution
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.