Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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))
 
Line 1: Line 1:
−
== 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 $\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}$. 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, Answer Key)
Preceded by:
Problem 38 		Followed by:
Problem 40
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_39&oldid=96782"