Solve the Equation

Question

Solve the Equation

 

x^-2/3 + x^-1/3 - 6 = 0

 

the solution set is ?

Answer 1

  X ^ ( -2/3) + X ^ ( -1/3) - 6 =0

 

 

    Let  X ^-1/3 = T

 

 

     T^2 + T - 6 =0

 

     ( T +3 ) ( T -2) =0

 

       T = -3     T =2

 

              X ^ ( -1/3) = 2

 

            [ X ^ ( -1/3) ] ^ ( -3) = 2 ^ ( -3)  

 

                   X = 1/8

 

              X ^( -1/3 )= -3

           

             [ X ^ ( -1/3) ] ^ -3 = -3 ^ ( -3)

 

                       X =- 1/27      

 

Answer 2

Hi Keith;

x^-2/3 + x^-1/3 - 6 = 0

This is no different than any other quadratic formula.  What is confusing you is the fractions within the exponentials.

(x^-1/3+3)(x^-1/3-2)=0

For the FOIL...

FIRST...(x^-1/3)(x^-1/3)...x^-2/3

Please remember that when multiplying exponentials, the exponentials must be added together.

OUTER...(x^-1/3)(-2)=-2x^-1/3

INNER...(3)(x^-1/3)=3x^-1/3

LAST...(3)(-2)=-6

0=x^-2/3-2x-^1/3+3x^-1/3-6

0=x^-2/3+x^1/3-6

0=(x^-1/3+3)(x^-1/3-2)

Either or both parenthetical equation(s) must equal zero...

0=x^-1/3+3

-3=x^-1/3

Let's raise both sides by -3...

-3^-3=(x^-1/3)^-3

A negative exponential means flipping x^-n into 1/xⁿ.  By multiplying a negative exponential by a negative exponential, we are flipping it again.

-1/3³=x

-1/27=x

0=x^-1/3-2

2=x^-1/3

2^-3=(x^-1/3)^-3

1/2³=x

1/8=x