Difference between revisions of "2007 AIME I Problems/Problem 8"
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Notice that if the roots of <math>Q_1(x)</math> and <math>Q_2(x)</math> are all distinct, then <math>P(x)</math> would have four distinct roots, which is a contradiction since it's cubic. Thus, <math>Q_1(x)</math> and <math>Q_2(x)</math> must share a root. Let this common value be <math>r.</math> | Notice that if the roots of <math>Q_1(x)</math> and <math>Q_2(x)</math> are all distinct, then <math>P(x)</math> would have four distinct roots, which is a contradiction since it's cubic. Thus, <math>Q_1(x)</math> and <math>Q_2(x)</math> must share a root. Let this common value be <math>r.</math> | ||
− | Thus, we see that we have \begin{align*} | + | Thus, we see that we have <math>\begin{align*} |
& r^2 + (k - 29)r - k = 0, \ | & r^2 + (k - 29)r - k = 0, \ | ||
− | & 2r^2 + (2k - 43)r + k = 0. \end{align*} Adding the two equations gives us <cmath>3r^2 + 3k - 72r = 0 \implies r = 0, 24 - k.</cmath> Now, we have two cases to consider. If <math>r = 0,</math> then we have that <math>Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.</math> On the other hand, if <math>r = 24 - k,</math> we see that <cmath>Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.</cmath> This can easily be checked to see that it does indeed work, and we're done! | + | & 2r^2 + (2k - 43)r + k = 0. \end{align*}</math> Adding the two equations gives us <cmath>3r^2 + 3k - 72r = 0 \implies r = 0, 24 - k.</cmath> Now, we have two cases to consider. If <math>r = 0,</math> then we have that <math>Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.</math> On the other hand, if <math>r = 24 - k,</math> we see that <cmath>Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.</cmath> This can easily be checked to see that it does indeed work, and we're done! |
~Ilikeapos | ~Ilikeapos |
Revision as of 19:47, 29 August 2020
Problem
The polynomial is cubic. What is the largest value of for which the polynomials and are both factors of ?
Contents
[hide]Solution
Solution 1
We can see that and must have a root in common for them to both be factors of the same cubic.
Let this root be .
We then know that is a root of , so .
We then know that is a root of so we get: or , so is the highest.
We can trivially check into the original equations to find that produces a root in common, so the answer is .
Solution 2
Again, let the common root be ; let the other two roots be and . We can write that and that .
Therefore, we can write four equations (and we have four variables), , , , and .
The first two equations show that . The last two equations show that . Solving these show that and that . Substituting back into the equations, we eventually find that .
Solution 3
Since and are both factors of , which is cubic, we know the other factors associated with each of and must be linear. Let , where and . Then we have that . Equating coefficients, we get the following system of equations:
Using equations and to make substitutions into equation , we see that the 's drop out and we're left with . Substituting this expression for into equation and solving, we see that must be .
~ anellipticcurveoverq
Solution 4
Notice that if the roots of and are all distinct, then would have four distinct roots, which is a contradiction since it's cubic. Thus, and must share a root. Let this common value be
Thus, we see that we have $
~Ilikeapos
Video Solution
https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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