Question 624828: x^(2/3)-3x^(1/3)-18=0 help solve x
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
Since you chose the "square-cubic-other-roots" category you must realize that those fractional exponents with denominators of 3 represent cube roots.
A straightforward approach to solving radical equations is to:- Isolate a radical
- Raise both sides of the equation to an appropriate power (in this case, to the 3rd power)
- Repeat these steps until all the radicals are gone.
- Solve the resulting equation.
- Check the solution(s) (required if you raised the equation to an even-numbered power)
But this is the hard way to solve this equation. Not only is cubing an equation a pain, but we would have to do it twice to get rid of both radicals! And on top of that the radical-less equation we end up with would be very difficult to solve.
Fortunately there is a much easier way to solve this equation. The exponent of 2/3 is exactly twice as large as the other exponent, 1/3. This means the equation is in what is called "quadratic form". ("Pure" quadratic equations as you should know, have exponents of 2 and 1. (Note that the 2 is twice the 1!)) Quadratic form equations can be solved using the similar methods to the ones used on quadratic equations (as you'll see shortly). So, as it turns out, this is more of a quadratic equation problem than a cube root problem.
Until you have done a few of these quadratic form problems, it can be a bit confusing. It can help to use a temporary variable. Set the temporary variable equal to the expression being raised to the lower power (in this case 1/3 power):
Let  . Then  Substituting these into your equation we get:
This is obviously a quadratic equation. One side is already zero, so to solve this the next step is to factor it (or use the quadratic formula). This factors easily:
From the Zero Product Property we know that one of these factors must be zero:
q - 6 = 0 or q + 3 = 0
Solving these we get:
q = 6 or q = -3
Of course we're not interested in values for q. We want solutions for x. So now we substitute back for q (it was a temporary variable after all):
 or 
To solve for x we just cube both equations:
 or 
x = 216 or x = -27
Since we did not raise both sides of the equation to an even-number power (3 is odd, of course), we are not required to check. You are welcome to do so.
P.S.- This may not have seemed easy. But believe me, this was much easier than cubing twice and then solving a 6th degree polynomial whose solutions are 216 and -27!!
- Once you get comfortable with these quadratic form equations you will not need a temporary variable. You will "see" how to go directly from
to
to
 or 
etc.
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