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

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From these equations, we can write that <math>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2</math>, and simplifying gives us <math>2s-5r-13=0</math> or <math>s = \frac{5r+13}{2}</math>.
 
From these equations, we can write that <math>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2</math>, and simplifying gives us <math>2s-5r-13=0</math> or <math>s = \frac{5r+13}{2}</math>.
  
−
We know move to the other two equations. We see that we can cancel a negative from both sides to get <math>rs(r+s) = b</math> and <math>(r+4)(s-3)(r+s+1)=b + 240</math>. Subtracting the first from the second equation gives us <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>. Expanding and simplifying, substituting <math>s = \frac{5r+13}{2}</math> and simplifying some more yields the simple quadratic <math>r^2 + 4r - 5 = 0</math>, so <math>r = -5, 1</math>. Then <math>s = -6, 9</math>.  
+
We now move to the other two equations. We see that we can cancel a negative from both sides to get <math>rs(r+s) = b</math> and <math>(r+4)(s-3)(r+s+1)=b + 240</math>. Subtracting the first from the second equation gives us <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>. Expanding and simplifying, substituting <math>s = \frac{5r+13}{2}</math> and simplifying some more yields the simple quadratic <math>r^2 + 4r - 5 = 0</math>, so <math>r = -5, 1</math>. Then <math>s = -6, 9</math>.  
  
 
Finally, we substitute back in to get <math>b = (-5)(-6)(-5-6) = -330</math> or <math>b = (1)(9)(1 + 9) = 90</math>. Then the answer is <math>|-330|+|90| = \boxed{420}</math>.
 
Finally, we substitute back in to get <math>b = (-5)(-6)(-5-6) = -330</math> or <math>b = (1)(9)(1 + 9) = 90</math>. Then the answer is <math>|-330|+|90| = \boxed{420}</math>.

Revision as of 22:29, 10 October 2016

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 $ax$ 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: \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] 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 $\boxed{420}$.

Solution 2

As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$.

From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \frac{5r+13}{2}$.

We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$.

Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \boxed{420}$.

See also

2014 AIME II (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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