Difference between revisions of "2007 iTest Problems/Problem 39"
(Created page with "== Problem == Let a and b be relatively prime positive integers such that a/b is the sum of the real solutions to the equation <math>\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+...") |
Rockmanex3 (talk | contribs) (Solution to Problem 39 (credit to grav123 and djmathman)) |
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− | == Problem == | + | ==Problem== |
Let a and b be relatively prime positive integers such that a/b is the sum of the real solutions to the equation <math>\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}</math>. Find <math>a+b</math>. | Let a and b be relatively prime positive integers such that a/b is the sum of the real solutions to the equation <math>\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}</math>. Find <math>a+b</math>. | ||
− | == Solution == | + | ==Solution== |
+ | |||
+ | Let <math>A = \sqrt[3]{3x-4}, B = \sqrt[3]{5x-6}, C = \sqrt[3]{x-2},</math> and <math>D = \sqrt[3]{7x-8}.</math> With these substitutions, we know that | ||
+ | <cmath>\begin{align*} | ||
+ | A+B &= C+D \\ | ||
+ | A^3 + B^3 &= C^3 + D^3. | ||
+ | \end{align*}</cmath> | ||
+ | We can factor the equation with the cubics and do some substitution. | ||
+ | <cmath>\begin{align*} | ||
+ | (A+B)(A^2 - AB + B^2) &= (C+D)(C^2 - CD + D^2) \\ | ||
+ | (A+B)((A+B)^2 - 3AB) &= (C+D)((C+D)^2 - 3CD) | ||
+ | \end{align*}</cmath> | ||
+ | Bringing all the terms into one side results in | ||
+ | <cmath>\begin{align*} | ||
+ | 0 &= (C+D)((C+D)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\ | ||
+ | 0 &= (A+B)((A+B)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\ | ||
+ | 0 &= (A+B)(3AB - 3CD) | ||
+ | \end{align*}</cmath> | ||
+ | By the Zero Product Property, either <math>A+B = 0</math> or <math>3AB-3CD = 0.</math> | ||
+ | |||
+ | <br> | ||
+ | If <math>A+B=0,</math> then <math>\sqrt[3]{3x-4} = -\sqrt[3]{5x-6}.</math> Cubing both sides yields <math>3x-4 = -5x+6,</math> so <math>x = \tfrac54.</math> | ||
+ | |||
+ | <br> | ||
+ | If <math>3AB-3CD = 0,</math> then <math>3\sqrt[3]{(3x-4)(5x-6)} = 3\sqrt[3]{(x-2)(7x-8)}.</math> Dividing both sides by three and cubing both sides yields <math>(3x-4)(5x-6) = (x-2)(7x-8).</math> Multiplying the binomials results in <math>15x^2 - 38x + 24 = 7x^2 - 22x + 16,</math> and rearranging and factoring leads to <math>8(x-1)^2 = 0.</math>. That means <math>x=1.</math> | ||
+ | |||
+ | <br> | ||
+ | Since both values of <math>x</math> satisfy the original equation, <math>\tfrac{a}{b} = \tfrac54 + 1 = \tfrac94,</math> so <math>a+b = \boxed{13}.</math> | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=38|num-a=40}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 21:27, 4 August 2018
Problem
Let a and b be relatively prime positive integers such that a/b is the sum of the real solutions to the equation . Find .
Solution
Let and With these substitutions, we know that We can factor the equation with the cubics and do some substitution. Bringing all the terms into one side results in By the Zero Product Property, either or
If then Cubing both sides yields so
If then Dividing both sides by three and cubing both sides yields Multiplying the binomials results in and rearranging and factoring leads to . That means
Since both values of satisfy the original equation, so
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 38 |
Followed by: Problem 40 | |
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