Question 623059
{{{6x^(2/3)-7x^(1/3) -24 = 0}}}
Equations where the exponent of the first (highest degree) term is twice the exponent of the middle term are equations of "quadratic form". And they can be solved in a similar way.<br>
It can help to use a temporary variable:
Let {{{q = x^(1/3)}}}
Then {{{q^2 = (x^(1/3))^2 = x^(2/3)}}}
Substituting these into our equation we get:
{{{6q^2-7q-24 = 0}}}
It now looks like a quadratic equation. We can go ahead an solve for q. Factoring we get:
{{{(2q+3)(3q-8) = 0}}}
From the Zero Product Property:
2q+3 = 0 or 3q-8 = 0
2q = -3 or 3q = 8
q = -3/2 or q = 8/3<br>
But we are not interested in a value for q. We are interested in x. So we now substitute back for x:
{{{x^(1/3) = -3/2}}} or {{{x^(1/3) = 8/3}}}
To solve for x we just cube both sides:
{{{(x^(1/3))^3 = (-3/2)^3}}} or {{{(x^(1/3))^3 = (8/3)^3}}}
{{{x = -27/8}}} or {{{x = 512/27}}}<br>
P.S. Once you have some practice with these quadratic form equations, you will not need a temporary variable. You will see how to factor
{{{6x^(2/3)-7x^(1/3) -24 = 0}}}
into
{{{(2x^(1/3)+3)(3x^(1/3)-8) = 0}}}
etc.