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4x^1/3 -2* x/x^2/3 = 7+ sqrt3(x)

to clarify, the exponents are fractions 

 Jun 29, 2022
 #1
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\(4x ^ { 1 \over 3} -2{x \over x^{2 \over 3}} = 7 + \sqrt {3x}\)

 

This is the correct equation right?

 Jun 29, 2022
 #2
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Right but what is x?

elroberto007  Jun 29, 2022
 #3
avatar+2448 
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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..... 

 Jun 29, 2022
edited by BuilderBoi  Jun 29, 2022

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