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Difference between revisions of "2014 AIME II Problems/Problem 5"
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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  + 240 = 12r^2 + 48r + 304 + 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 + 147 = 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  + 240 = 12r^2 + 48r + 304 + 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 + 147 = 0</cmath>
 +
 
 +
<cmath>r^2 - s^2 + 4r + 3s + 49 = 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 <math>r = -5</math> and <math>r = 1</math>, corresponding to <math>s = -6</math> and <math>s = 9</math>, and <math>|b| = 330, 90</math>, for an answer of <math>\boxed{420}</math>.

Revision as of 11:45, 31 March 2014

Problem 5

Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.


Solution

Let r, s, -r-s be the roots of p(x) (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for s. Also, \[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\]

Set up a similar equation for s:

\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0\].

Simplifying and adding the equations gives \[3r^2 - 3s^2 + 12r + 9s + 147 = 0\]

\[r^2 - s^2 + 4r + 3s + 49 = 0 (*)\]

Now, let's deal with the a*x. Equating the a in both equations (per Vieta) \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] which eventually simplifies to

\[s = \frac{13 + 5r}{2}.\]

Substitution into (*) should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\boxed{420}$.

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