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

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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>.
  
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==Solution 2==
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==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 (co-efficient of <math>x^2</math> term is <math>0</math>).  
  

Revision as of 14:50, 31 December 2023

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 $s = \tfrac 12 (5r+13)$. The product of the roots of $p(x)$ differs from that of $q(x)$ by $240$, so\[(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240\]from which $r^2+4r-5 =0$, with roots $r=1$ and $r=-5$.

If $r = 1$, then the roots of $p(x)$ are $r=1$, $s=9$, and $-(r+s)=-10$, and $b=-rst=90$.

If $r = -5$, then the roots of $p(x)$ are $r=-5$, $s=-6$, and $-(r+s)=11$, and $b=-rst=-330$.

Therefore the requested sum is $|- 330| + |90| = \boxed{420}$.

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 (co-efficient 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 3

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
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