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Difference between revisions of "2014 AIME II Problems/Problem 5"

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<cmath>r^2 - s^2 + 4r + 3s + 49 = 0 (*)</cmath>
 
<cmath>r^2 - s^2 + 4r + 3s + 49 = 0 (*)</cmath>
  
−
Now, let's deal with the <math>a*x</math>. Equating the a in both equations (per Vieta)
+
Now, let's deal with the <math>a*x</math> terms. Plugging the roots <math>r</math>, <math>s</math>, and <math>-r-s</math> into <math>p(x)</math> yields a long polynomial, and plugging the roots <math>r+4</math>, <math>s-3</math>, and <math>-1-r-s</math> into <math>q(x) yields another long polynomial. Equating the coefficients of x in both polynomials:
 
<cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath>
 
<cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath>
 
which eventually simplifies to
 
which eventually simplifies to
Line 22: Line 22:
 
<cmath>s = \frac{13 + 5r}{2}.</cmath>
 
<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>.
+
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}$.
  
 
== See also ==
 
== See also ==

Revision as of 16:02, 9 February 2015

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 $r$, $s$, and $-r-s$ 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 $a*x$ terms. Plugging the roots $r$, $s$, and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$, $s-3$, and $-1-r-s$ into $q(x) yields another long polynomial. Equating the coefficients of x in both polynomials: <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$ (Error compiling LaTeX. Unknown error_msg)r = -5$and$r = 1$, corresponding to$s = -6$and$s = 9$, and$|b| = 330, 90$, for an answer of$\boxed{420}$.

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

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