How do I solve an equation with fractional exponents?

Question

I'm having trouble with solving the equation
x^(1/2) - 3x^(1/3) = 3x^(1/6) - 9

I tried substituting for different variables, but I always end up with 3 terms that won't combine. ie. w^2 = x^(1/6) w = x^(1/3), but what is x^(1/2)

Answer 1

x^(1/2)=√x. Similarly, x^(1/n) is the nth root of x. The way you solve this equation is to make the substitution w=x^(1/6), thus x^(1/3)=w² andx^(1/2)=w³. Thus you get w³-3w²-3w+9=0. You find that this factors into (w-3)(w²-3)=0, which then becomes (w-3)(w-√3)(w+√3)=0, so the roots of this equation are w=3, w=√3, and w=-√3. Then you simply find x - since x^(1/6)=w, w^6=x, so x=729 or x=27 (√3 and -√3 give the same value of x).

Answer 2

Try w^6 = x^1/6, w^3 = x^1/3, w^2 = x^1/2. You had the right idea, just missed the execution for some reason.

Answer 3

enable y = x^(a million/6) (only define a sparkling variable y) it follows that x^(a million/3) = x^(2/6) = (x^(a million/3))^2 = y^2. So your equation is 2x^(a million/3)+5x^(a million/6)-3 = 0 exchange making use of the values above to get 2y^2 + 5y - 3 = 0. resolve this equation with the quadratic equation: y = (-5 +/- sqrt(25-4*2*(-3)))/(2*2) = (-5+/-7)/4 y = -3 or a million/2 on account that y = x^(a million/6), x = y^6, so x = 729 or a million/sixty 4

Answer 4

The easist way is to let y = x^(1/6)
so x^(1/2) = x^(3/6) = y^3
and x^(1/3) = x^(2/6) = y^2
then off you go with your nice cubic equation.

Answer 5

Let x = w^6, then

w^(6/2) - 3w^(6/3) = 3w^(6/6) - 9
w^3 - 3w^2 = 3w - 9
=> w^3 - 3w^2 - 3w + 9 = 0

The solutions are:

w1 = +1.73205
w2 = -1.73205
w3 = +3

From there you can find the solutions for x.

I hope this helps.