
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 3rd, 2019, 07:47 PM  #1 
Newbie Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0  Solving roots equation neatly
I am to find all real solutions for this problem. $2\sqrt[3]{x^2}\sqrt[3]{x}=1$ According to my textbook, the solutions are {$\frac{1}{8}$, 1}. However, I keep getting {$\frac{1}{2}$, 1} when I try to solve it more practically. I went on Symbolab to see how they do it and I learned if I use the exponent property $a^{n}=(\sqrt[m]{a})^{n\times m}$, I get the correct solution set. My issue with this, though, is that I'm not going to be able to remember that exponent property well and the work after that gets too tedious for this kind of problem. Here's one of my attempts of solving this problem to give you an idea of what I've been doing. $$2\sqrt[3]{x^2}\sqrt[3]{x}=1$$ $$=2x^{\frac{2}{3}}x^{\frac{1}{3}}=1$$ $$=(2x^{\frac{2}{3}}x^{\frac{1}{3}})^3=1^3$$ $$=2x^2x1=0$$ $$=x(2x+1)1(2x+1)=0$$ $$=(x1)(2x+1)=0$$ $$x={\frac{1}{2}, 1}$$ Of course that's a flawed solution, but can anyone help me find a more elegant approach to solving this problem? Please use TeX commands 
January 3rd, 2019, 07:53 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
$u = \sqrt[3]{x}$ $2u^2  u  1 = 0$ $(2u+1)(u1) = 0$ $u = \dfrac 1 2,~u=1$ $x = u^3$ $x = \dfrac 1 8,~x=1$ 
January 3rd, 2019, 07:54 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra 
$$(a+b)^3 \ne a^3 + b^3$$ Specifically, $$(2x^\tfrac23  x^\tfrac13)^3 \ne 2x^2x$$ 
January 3rd, 2019, 08:02 PM  #4 
Newbie Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0 
Wow. I can't believe I made that mistake. I'll try to actually expand it and see where it goes from there.

January 3rd, 2019, 08:15 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
Letting $x = y^3\!$, so that $y = \sqrt[3]{x}$, gives $2y^2  y  1 = 0$. Hence $y = 1/2$ or $1$, and so $x = 1/8$ or $1$.


Tags 
equation, equations, exponents, neatly, polynomials, quadratics, roots, solving 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Roots of the equation x2 –3x –5 = 0  Nosidak  Algebra  7  January 24th, 2012 01:20 AM 
Solving equation in rational roots  rhseiki  Number Theory  18  May 18th, 2011 02:08 PM 
Solving real roots polynomial equation  RKJCHENNAI  Abstract Algebra  5  March 5th, 2010 05:12 PM 
roots and equation  damgam  Algebra  1  December 22nd, 2008 02:28 PM 
Solving roots  Zets  Algebra  2  October 8th, 2008 05:05 AM 