Difference between revisions of "2007 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
− | The polynomial <math>P(x)</math> is cubic. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both | + | The [[polynomial]] <math>P(x)</math> is [[cubic polynomial | cubic]]. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both [[factor]]s of <math>P(x)</math>? |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
− | + | We can see that <math>Q_1</math> and <math>Q_2</math> must have a [[root]] in common for them to both be [[factor]]s of the same cubic. | |
− | + | Let this root be <math>a</math>. | |
− | < | + | We then know that <math>a</math> is a root of |
+ | <math> | ||
+ | Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 | ||
+ | </math> | ||
+ | , so <math>x = \frac{-k}{5}</math>. | ||
− | + | We then know that <math>\frac{-k}{5}</math> is a root of <math>Q_{1}</math> so we get: | |
+ | <math> | ||
+ | \frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k | ||
+ | </math> | ||
+ | or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest. | ||
+ | |||
+ | We can trivially check into the original equations to find that <math>k=30</math> produces a root in common, so the answer is <math>030</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Again, let the common root be <math>a</math>; let the other two roots be <math>m</math> and <math>n</math>. We can write that <math>(x - a)(x - m) = x^2 + (k - 29)x - k</math> and that <math>2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)</math>. | ||
+ | |||
+ | Therefore, we can write four equations (and we have four [[variable]]s), <math>a + m = 29 - k</math>, <math>a + n = \frac{43}{2} - k</math>, <math>am = -k</math>, and <math>an = \frac{k}{2}</math>. | ||
+ | |||
+ | The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>. Solving these show that <math>m = 5</math> and that <math>n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = 30</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Since <math>Q_1(x)</math> and <math>Q_2(x)</math> are both factors of <math>P(x)</math>, which is cubic, we know the other factors associated with each of <math>Q_1(x)</math> and <math>Q_2(x)</math> must be linear. Let <math>Q_1(x)R(x) = Q_2(x)S(x) = P(x)</math>, where <math>R(x) = ax + b</math> and <math>S(x) = cx + d</math>. Then we have that <math>((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)</math>. Equating coefficients, we get the following system of equations: | ||
+ | |||
+ | <cmath>\begin{align} | ||
+ | a = 2c \\ | ||
+ | b = -d \\ | ||
+ | 2c(k - 29) - d = c(2k - 43) + 2d \\ | ||
+ | -d(k - 29) - 2ck = d(2k - 43) + ck | ||
+ | \end{align}</cmath> | ||
+ | |||
+ | Using equations <math>(1)</math> and <math>(2)</math> to make substitutions into equation <math>(3)</math>, we see that the <math>k</math>'s drop out and we're left with <math>d = -5c</math>. Substituting this expression for <math>d</math> into equation <math>(4)</math> and solving, we see that <math>k</math> must be <math>\boxed {30}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | 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 | ||
+ | <cmath>r^2 + (k - 29)r - k = 0,</cmath> | ||
+ | <cmath>2r^2 + (2k - 43)r + k = 0.</cmath> 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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s | ||
== See also == | == See also == | ||
{{AIME box|year=2007|n=I|num-b=7|num-a=9}} | {{AIME box|year=2007|n=I|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:48, 29 August 2020
Problem
The polynomial is cubic. What is the largest value of for which the polynomials and are both factors of ?
Contents
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 Adding the two equations gives us Now, we have two cases to consider. If then we have that On the other hand, if we see that This can easily be checked to see that it does indeed work, and we're done!
~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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