Difference between revisions of "2014 AIME II Problems/Problem 5"
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | ||
+ | ==Hint== | ||
+ | <cmath>\textbf{USE VIÈTE'S FORMULAE!!!}</cmath> | ||
==Solution 1== | ==Solution 1== | ||
Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, | Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, |
Revision as of 15:38, 24 August 2021
Problem
Real numbers and are roots of , and and are roots of . Find the sum of all possible values of .
Hint
Solution 1
Let , , and be the roots of (per Vieta's). Then and similarly for . Also,
Set up a similar equation for :
Simplifying and adding the equations gives
Now, let's deal with the terms. Plugging the roots , , and into yields a long polynomial, and plugging the roots , , and into yields another long polynomial. Equating the coefficients of x in both polynomials: which eventually simplifies to
Substitution into (*) should give and , corresponding to and , and , for an answer of .
Solution 2
The roots of are , , and since they sum to by Vieta's Formula (co-efficient of term is ).
Similarly, the roots of are , , and , as they too sum to .
Then:
and from and
and from .
From these equations, we can write that and simplifying gives
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get Subtracting the first equation from the second equation gives us .
Expanding, simplifying, substituting , and simplifying some more yields the simple quadratic , so . Then .
Finally, we substitute back into to get , or .
The answer is .
Solution 3
By Vieta's, we know that the sum of roots of is . Therefore, the roots of are . By similar reasoning, the roots of are . Thus, and .
Since and have the same coefficient for , we can go ahead and match those up to get
At this point, we can go ahead and compare the constant term in and . Doing so is certainly valid, but we can actually do this another way. Notice that . Therefore, . If we plug that into our expression, we get that This tells us that or . Since is the product of the roots, we have that the two possibilities are and . Adding the absolute values of these gives us .
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.