A complex algebra problem.?

Question

I just can't seem to figure out this problem because I forgot many of the laws, and I can't find them in my notes from previous math courses:

6x^(3/4)=7x^(1/4)-2x^(- 1/4)

I attempted to use the a2+b2=c2, but I'm not sure how the exponents would play out.

Answer 1

6x^(3/4)=7x^(1/4)-2x^(- 1/4)

Let u = x^(1/4)

6u^3 = 7u - (2/u)

6u^4 = 7u^2 - 2

6u^4 - 7u^2 + 2 = 0

(3u^2 - 2)(2u^2 - 1) = 0

u^2 = 2/3 or 1/2

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

x = 4/9 or 1/4

Answer 2

im not sure. maybe you can try turning the fractions to decimals and see if that helps.

Answer 3

First multiply through by x^(1/4), which gives:

6x = 7sqrt(x) - 2

6x - 7sqrt(x) + 2 = 0

(3sqrt(x) - 2)(2sqrt(x) - 1) = 0

sqrt(x) = 2/3 => x = 4/9  

OR

sqrt(x) = 1/2 => x = 1/4

You can't really use a^2 + b^2 = c^2, since this only holds when a, b, and c are the sides of a right triangle.