Question

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

Answer 1

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


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

Answer 2

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?