Question

How do you find the x intercepts of 3x^(5/3) - 4x^(7/3)

Answer 1

Set the whole expression = to 0 and solve for x.

3x^(5/3) - 4x^(7/3) = 0.  Factor out x^(5/3):     x^(5/3) [3 - 4x^(2/3)] = 0

Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.

In the latter case, 4x^(2/3) = 3. 

To solve this:  mult. both sides by x^(-2/3).  Then we have

4x^(2/3)x^(-2/3) = 3x^(-2/3),            or 4 = 3x^(-2/3).  It'd be easier to work with this if we rewrote it as

4           3
---  = --------------------
1            x^(+2/3)

Then 

4
---  = x^(-2/3).  Then, x^(2/3) = (3/4), and x = (3/4)^(3/2).  According to my     3                      calculator, that comes out to x = 0.65 (approx.)

Check this result!  subst. 0.65 for x in the given equation.  Is the equation then true?

My method here was a bit roundabout, and longer than it should have been.  Can you think of a more elegant (and shorter) solution?

Answer 2

to the risk of sounding redundant, as Altavistard already pointed out above

$\bf 3x^{\frac{5}{3}}-4x^{\frac{7}{3}}=0\implies \stackrel{common~factor}{x^{\frac{5}{3}}}(3-4x^{\frac{2}{3}})=0\\\\ -------------------------------\\\\ x^{\frac{5}{3}}=0\implies \boxed{x=0}\\\\ -------------------------------\\\\ 3-4x^{\frac{2}{3}}=0\implies 3=4x^{\frac{2}{3}}\implies (3)^3=\left(4x^{\frac{2}{3}}\right)^3 \\\\\\ 27=64x^2\implies \cfrac{27}{64}=x^2\implies \sqrt{\cfrac{27}{64}}=x\implies \boxed{\cfrac{3\sqrt{3}}{4\sqrt{4}}=x} \\\\\\ 0.64951905283832898507\approx x$