Difference between revisions of "2014 AIME II Problems/Problem 5"
(Created page with "==Solution== Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, <cmath>q(r+4) = (r+4)^3 + a(r+4) + b = 12r^2 + 48r +...") |
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| + | ==Problem 5== | ||
| + | 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>. | ||
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==Solution== | ==Solution== | ||
Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, <cmath>q(r+4) = (r+4)^3 + a(r+4) + b = 12r^2 + 48r + 64 + 4a = 0</cmath> Set up a similar equation for s: <cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0</cmath>. Simplifying and adding the equations gives <cmath>3r^2 - 3s^2 + 12r + 9s + 87 = 0 (*)</cmath> Now, let's deal with the a*x. Equating the a in both equations (per Vieta) <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)</cmath>, which eventually simplifies to <cmath>s = \frac{13 + 5r}{2}</cmath> Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of <math>\boxed{420}</math>. | Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, <cmath>q(r+4) = (r+4)^3 + a(r+4) + b = 12r^2 + 48r + 64 + 4a = 0</cmath> Set up a similar equation for s: <cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0</cmath>. Simplifying and adding the equations gives <cmath>3r^2 - 3s^2 + 12r + 9s + 87 = 0 (*)</cmath> Now, let's deal with the a*x. Equating the a in both equations (per Vieta) <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)</cmath>, which eventually simplifies to <cmath>s = \frac{13 + 5r}{2}</cmath> Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of <math>\boxed{420}</math>. | ||
Revision as of 21:00, 28 March 2014
Problem 5
Real numbers 
and 
are roots of 
, and 
and 
are roots of 
. Find the sum of all possible values of 
.
Solution
Let r, s, -r-s be the roots of p(x) (per Vieta's). Then 
and similarly for s. Also, ![\[q(r+4) = (r+4)^3 + a(r+4) + b = 12r^2 + 48r + 64 + 4a = 0\]](http://latex.artofproblemsolving.com/9/a/2/9a279c546a308a06b75eb62b1234b422a109ef51.png)
Set up a similar equation for s: ![\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0\]](http://latex.artofproblemsolving.com/0/3/4/034672706f6a607dd7cd2d61f24a6e64cea3a61a.png)
. Simplifying and adding the equations gives ![\[3r^2 - 3s^2 + 12r + 9s + 87 = 0 (*)\]](http://latex.artofproblemsolving.com/d/0/3/d03e950442c1933c64ffd0b60547cdc307643d4d.png)
Now, let's deal with the a*x. Equating the a in both equations (per Vieta) ![\[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)\]](http://latex.artofproblemsolving.com/5/1/0/51078910683e93bfd6c2dad4a38ec3987a6b3521.png)
, which eventually simplifies to ![\[s = \frac{13 + 5r}{2}\]](http://latex.artofproblemsolving.com/0/1/f/01feb9ac5c07ed2951e38eb46807fc301f8964f4.png)
Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of 
.
