Question 720553
{{{x^(2/3)-13x^(1/3)-30=0}}}
Since the exponent of {{{x^(2/3)}}} is exactly twice the exponent of {{{x^(1/3)}}}, this equation is in what is called quadratic form. And these quadratic form equations can be solved in much the same way as regular quadratic equations.<br>
To see the "quadratic-ness" of this equation it can help to use a temporary variable. Set it to the base and lower exponent. So:
Let {{{q = x^(1/3)}}}. 
Then {{{q^2 = (x^(1/3))^2 = x^(2/3)}}}.
Substituting these into our equation we get:
{{{q^2-13q-30=0}}}
This is obviously a quadratic equation. It's already got a zero on one side. And it factors pretty easily:
(q-10)(q-3) = 0
From the Zero Product Property:
q-10 = 0 or q-3 = 0
Solving these:
q = 10 or q = 3<br>
Of course we're not interested in solutions for q. We're interested in solutions for x. So we substitute back:
{{{x^(1/3) = 10}}} or {{{x^(1/3) = 3}}}
We have a little more work to do to solve for x. The quick way is to cube both sides:
{{{(x^(1/3))^3 = (10)^3}}} or {{{(x^(1/3))^3 = (3)^3}}}
which simplifies to:
x = 1000 or x = 27<br>
P.S. Once you have done a few of these quadratic form equations you will no longer need a temporary variable. You will learn to see how to go directly from:
{{{x^(2/3)-13x^(1/3)-30=0}}}
to
{{{(x^(1/3)-10)(x^(1/3)-3)=0}}}
to
{{{x^(1/3)-10 = 0}}} or {{{x^(1/3)-3=0}}}
etc.<br>
P.S. Please put fractional exponents in parentheses.