+0

# Solve for x:

+1
36
3

4x^1/3 -2* x/x^2/3 = 7+ sqrt3(x)

to clarify, the exponents are fractions

Jun 29, 2022

#1
+1

$$4x ^ { 1 \over 3} -2{x \over x^{2 \over 3}} = 7 + \sqrt {3x}$$

This is the correct equation right?

Jun 29, 2022
#2
+1

Right but what is x?

elroberto007  Jun 29, 2022
#3
+1

Yes, I'll solve it now, I just wanted to make sure I had the right question.

Anyways, we have: $$4x^{1 \over 3} - 2{x \over x^{2\over 3}} = 7 + \sqrt {3x}$$

First, note that $${x \over {x ^ {2 \over 3}}} = {x^{3 \over 3} \over x^{2 \over 3}} = x^{1 \over 3}$$

This means we have: $$4x^{1 \over 3} - 2{x^{1\over 3}} = 7 + \sqrt {3x}$$

Subtracting like terms gives us: $$2x^{1 \over 3} = 7 + \sqrt{3x}$$

To solve, let $$y = x^{1 \over 3}$$. This gives $$2y = 7 + \sqrt{3y^3}$$

We can convert this to $$0=-3y^3+ 4y^2-28y+49$$.

Solving for y gives us $$y \approx 1.65543$$, meaning $$x = \sqrty \approx1.18296$$

However, plugging this in doesn't work, meaning the solution is imaginary....

According to WA, its $$\approx -0.161 + 3.137i$$.

However, you should probably plug it in the original equation to see if it works.....

Jun 29, 2022
edited by BuilderBoi  Jun 29, 2022