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}+...")
 
(Solution to Problem 39 (credit to grav123 and djmathman))
 
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== Problem ==
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==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 ==
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==Solution==
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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
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<cmath>\begin{align*}
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A+B &= C+D \\
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A^3 + B^3 &= C^3 + D^3.
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\end{align*}</cmath>
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We can factor the equation with the cubics and do some substitution.
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<cmath>\begin{align*}
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(A+B)(A^2 - AB + B^2) &= (C+D)(C^2 - CD + D^2) \\
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(A+B)((A+B)^2 - 3AB) &= (C+D)((C+D)^2 - 3CD)
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\end{align*}</cmath>
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Bringing all the terms into one side results in
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<cmath>\begin{align*}
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0 &= (C+D)((C+D)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\
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0 &= (A+B)((A+B)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\
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0 &= (A+B)(3AB - 3CD)
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\end{align*}</cmath>
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By the Zero Product Property, either <math>A+B = 0</math> or <math>3AB-3CD = 0.</math>
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<br>
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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>
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<br>
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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>
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<br>
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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>
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==See Also==
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{{iTest box|year=2007|num-b=38|num-a=40}}
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[[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 $\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}$. Find $a+b$.

Solution

Let $A = \sqrt[3]{3x-4}, B = \sqrt[3]{5x-6}, C = \sqrt[3]{x-2},$ and $D = \sqrt[3]{7x-8}.$ With these substitutions, we know that \begin{align*} A+B &= C+D \\ A^3 + B^3 &= C^3 + D^3. \end{align*} We can factor the equation with the cubics and do some substitution. \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*} Bringing all the terms into one side results in \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*} By the Zero Product Property, either $A+B = 0$ or $3AB-3CD = 0.$


If $A+B=0,$ then $\sqrt[3]{3x-4} = -\sqrt[3]{5x-6}.$ Cubing both sides yields $3x-4 = -5x+6,$ so $x = \tfrac54.$


If $3AB-3CD = 0,$ then $3\sqrt[3]{(3x-4)(5x-6)} = 3\sqrt[3]{(x-2)(7x-8)}.$ Dividing both sides by three and cubing both sides yields $(3x-4)(5x-6) = (x-2)(7x-8).$ Multiplying the binomials results in $15x^2 - 38x + 24 = 7x^2 - 22x + 16,$ and rearranging and factoring leads to $8(x-1)^2 = 0.$. That means $x=1.$


Since both values of $x$ satisfy the original equation, $\tfrac{a}{b} = \tfrac54 + 1 = \tfrac94,$ so $a+b = \boxed{13}.$

See Also

2007 iTest (Problems)
Preceded by:
Problem 38
Followed by:
Problem 40
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