4x^1/3 -2* x/x^2/3 = 7+ sqrt3(x)

to clarify, the exponents are fractions

elroberto007 Jun 29, 2022

#1**0 **

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

This is the correct equation right?

BuilderBoi Jun 29, 2022

#3**0 **

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 \(x = \sqrt[3]y \approx1.18296\).

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.....

BuilderBoi Jun 29, 2022