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

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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>.
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==Solution 3==
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By Vieta's, we know that the sum of roots of <math>p(x)</math> is <math>0</math>. Therefore,
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the roots of <math>p</math> are <math>r, s, -r-s</math>. By similar reasoning, the roots of <math>q(x)</math>
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are <math>r + 4, s - 3, -r - s - 1</math>. Thus, <math>p(x) = (x - r)(x - s)(x + r + s)</math>
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and <math>q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)</math>.
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Since <math>p(x)</math> and <math>q(x)</math> have the same coefficient for <math>x</math>, we can go ahead
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and match those up to get
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<cmath>\begin{align*}
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    rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\
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    0 &= -13 - 5r + 2s \\
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    s &= \frac{5r + 13}{2}
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\end{align*}</cmath>
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At this point, we can go ahead and compare the constant term in <math>p(x)</math> and
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<math>q(x)</math>. Doing so is certainly valid, but we can actually do this another way. Notice that <math>p(s) = 0</math>. Therefore, <math>q(s) = 240</math>. If we plug that into
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our expression, we get that
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<cmath>\begin{align*}
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    q(s)  &= 3(s - r - 4)(r + 2s + 1) \\
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    240 &= 3(s - r - 4)(r + 2s + 1) \\
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    240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\
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    80 &= (3r + 5)(3r + 7) \\
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    0 &= r^2 + 4r - 5
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\end{align*}</cmath>
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This tells us that <math>(r, s) = (1, 9)</math> or <math>(-5, -6)</math>. Since <math>-b</math> is the product of the roots, we have that the two possibilities are <math>1 \cdot 9 \cdot (-10) = -90</math>
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and <math>(-5)(-6)11 = 330</math>. Adding the absolute values of these gives us
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<math>\boxed{420}</math>.
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== See also ==
 
== See also ==

Revision as of 23:05, 19 February 2021

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

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

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 (ProblemsAnswer KeyResources)
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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