Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Solution 1)
 
(16 intermediate revisions by 9 users not shown)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
 +
Because the coefficient of <math>x^2</math> in both <math>p(x)</math> and <math>q(x)</math> is 0, the remaining root of <math>p(x)</math> is <math>-(r+s)</math>, and the remaining root of <math>q(x)</math> is <math>-(r+s+1)</math>. The coefficients of <math>x</math> in <math>p(x)</math> and <math>q(x)</math> are both equal to <math>a</math>, and equating the two coefficients gives
 +
<cmath>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 </cmath>from which <math>s = \tfrac 12 (5r+13)</math>.
 +
 +
==Solution 2==
 
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,
 
<cmath>q(r+4) = (r+4)^3 + a(r+4) + b  + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath>
 
<cmath>q(r+4) = (r+4)^3 + a(r+4) + b  + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath>
Line 11: Line 15:
  
 
Simplifying and adding the equations gives
 
Simplifying and adding the equations gives
−
<cmath>3r^2 - 3s^2 + 12r + 9s + 147 = 0</cmath>
+
<cmath>\begin{align}\tag{*}
−
 
+
r^2 - s^2 + 4r + 3s + 49 &= 0
−
<cmath>r^2 - s^2 + 4r + 3s + 49 = 0 (*)</cmath>
+
\end{align}</cmath>
−
 
+
Now, let's deal with the <math>ax</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)</math> yields another long polynomial. Equating the coefficients of <math>x</math> in both polynomials, we get:
−
Now, let's deal with the <math>ax</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)</math> 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
−
 
 
<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}</math>.
  
−
==Solution 2==
+
==Solution 3==
−
The roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math> since they sum to <math>0</math> by Vieta's Formula (co-efficient of <math>x^2</math> term is <math>0</math>).  
+
The roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math> since they sum to <math>0</math> by Vieta's Formula (coefficient of <math>x^2</math> term is <math>0</math>).  
  
 
Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>, as they too sum to <math>0</math>.  
 
Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>, as they too sum to <math>0</math>.  
Line 42: Line 43:
  
 
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  
 
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  
−
<cmath>rs(r+s) = b</cmath> and
+
<cmath>rs(r+s) = b</cmath>  
−
<cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation from the second equation yields <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>.  
+
<cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation 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>.  
+
Expanding, 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 into <math>b=rs(r+s)</math> to get <math>b = (-5)(-6)(-5-6) = -330</math>, or <math>b = (1)(9)(1 + 9) = 90</math>.  
  
−
==Solution 3==
+
The answer is <math>|-330|+|90| = \boxed{420}</math>.
 +
 
 +
==Solution 4==
  
 
By Vieta's, we know that the sum of roots of <math>p(x)</math> is <math>0</math>. Therefore,
 
By Vieta's, we know that the sum of roots of <math>p(x)</math> is <math>0</math>. Therefore,
Line 77: Line 80:
 
and <math>(-5)(-6)11 = 330</math>. Adding the absolute values of these gives us
 
and <math>(-5)(-6)11 = 330</math>. Adding the absolute values of these gives us
 
<math>\boxed{420}</math>.
 
<math>\boxed{420}</math>.
−
 
  
 
== See also ==
 
== See also ==

Latest revision as of 21:07, 18 August 2024

Problem

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 1

Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$, and the remaining root of $q(x)$ is $-(r+s+1)$. The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$, and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\]from which $s = \tfrac 12 (5r+13)$.

Solution 2

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 \begin{align}\tag{*} r^2 - s^2 + 4r + 3s + 49 &= 0 \end{align} 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, we get: \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] 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 3

The roots of $p(x)$ are $r$, $s$, and $-r-s$ since they sum to $0$ by Vieta's Formula (coefficient of $x^2$ term is $0$).

Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$, as they too sum to $0$.

Then:

$a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and

$a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$.

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


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 \[rs(r+s) = b\] \[(r+4)(s-3)(r+s+1)=b + 240.\] Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$.

Expanding, 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 into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$, or $b = (1)(9)(1 + 9) = 90$.

The answer is $|-330|+|90| = \boxed{420}$.

Solution 4

By Vieta's, we know that the sum of roots of $p(x)$ is $0$. Therefore, the roots of $p$ are $r, s, -r-s$. By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$. Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$.

Since $p(x)$ and $q(x)$ have the same coefficient for $x$, we can go ahead and match those up to get \begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\ 0 &= -13 - 5r + 2s \\ s &= \frac{5r + 13}{2} \end{align*}

At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$. Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$. Therefore, $q(s) = 240$. If we plug that into our expression, we get that \begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\ 80 &= (3r + 5)(3r + 7) \\ 0 &= r^2 + 4r - 5 \end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$. Since $-b$ is the product of the roots, we have that the two possibilities are $1 \cdot 9 \cdot (-10) = -90$ and $(-5)(-6)11 = 330$. Adding the absolute values of these gives us $\boxed{420}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_5&oldid=225285"