SOLUTION: Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.) 6x^2/3 − 7x^1/3 &#872

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.) 6x^2/3 − 7x^1/3 &#872      Log On


   

Question 623059: Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.)

6x^2/3 − 7x^1/3 − 24 = 0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
6x%5E%282%2F3%29-7x%5E%281%2F3%29+-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.

It can help to use a temporary variable:
Let q+=+x%5E%281%2F3%29
Then q%5E2+=+%28x%5E%281%2F3%29%29%5E2+=+x%5E%282%2F3%29
Substituting these into our equation we get:
6q%5E2-7q-24+=+0
It now looks like a quadratic equation. We can go ahead an solve for q. Factoring we get:
%282q%2B3%29%283q-8%29+=+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

But we are not interested in a value for q. We are interested in x. So we now substitute back for x:
x%5E%281%2F3%29+=+-3%2F2 or x%5E%281%2F3%29+=+8%2F3
To solve for x we just cube both sides:
%28x%5E%281%2F3%29%29%5E3+=+%28-3%2F2%29%5E3 or %28x%5E%281%2F3%29%29%5E3+=+%288%2F3%29%5E3
x+=+-27%2F8 or x+=+512%2F27

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%5E%282%2F3%29-7x%5E%281%2F3%29+-24+=+0
into
%282x%5E%281%2F3%29%2B3%29%283x%5E%281%2F3%29-8%29+=+0
etc.