Question 818015
{{{x^(2/3) + x^(1/3) - 2 = 0}}}
Since the exponent of the first term, 2/3, is exactly twice the exponent in the middle, 1/3, this equation is in what is known as quadratic form. Quadratic form equations have the same underlying structure as a "pure" quadratic equation and can be solved using the same techniques.<br>
To see this more easily, it can be helpful to use a temporary variable:
Let {{{q = x^(1/3)}}}. This makes {{{q^2 = (x^(1/3))^2 = x^(2/3)}}}.
Substituting these in to the equation we get:
{{{q^2+q-2=0}}}
This is obviously a quadratic equation. It will factor:
{{{(q+2)(q-1)=0}}}
From the Zero Product Property:
q+2 = 0 or q-1 = 0
Solving these:
q = -2 or q = 1<br>
Of course we are not interested in solutions for our made-up variable q. So we replace the q's:
{{{x^(1/3) = -2}}} or {{{x^(1/3) = 1}}}
To solve these we just cube both sides of both equations:
{{{x = -8}}} or {{{x = 1}}}<br>
P.S. After doing several of these quadratic form equations you will no longer need the temporary variable. You will start seeing how to go directly from:
{{{x^(2/3) + x^(1/3) - 2 = 0}}}
to
{{{(x^(1/3)+2)(x^(1/3)-1) = 0}}}
to
{{{x^(1/3)+2 = 0}}} or {{{x^(1/3)-1 = 0}}}
etc.