What is the answer to (x^3+x)/(3x^2-4)x=2 ?

The answer to (x^3+x)/(3x^2-4)x=2 is

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

\frac{x ^{3} + x}{3 x ^{2} - 4} x = 2

x = \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i

Solution steps

\frac{x ^{3} + x}{3 x ^{2} - 4} x = 2

Simplify

\frac{x ^{3} + x}{3 x ^{2} - 4} x : \frac{x ( x ^{3} + x )}{3 x ^{2} - 4}

\frac{x ( x ^{3} + x )}{3 x ^{2} - 4} = 2

Multiply both sides by

3 x^{2} - 4

x (x^{3} + x) = 2 (3 x^{2} - 4)

Solve

x (x^{3} + x) = 2 (3 x^{2} - 4) : x = \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i

x = \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = \frac{5 + 4 2}{2} - \frac{7}{2 5 + 4 2} i, x = - \frac{5 + 4 2}{2} + \frac{7}{2 5 + 4 2} i

2 x - 3 = \frac{2}{x}

- 3 = - \frac{2}{u + 2}

(x + \frac{1}{x})^{5} = 3^{5}

8 = - \frac{4}{v}

\frac{2}{x} + 2 x = 5

What is the answer to (x^3+x)/(3x^2-4)x=2 ?
The answer to (x^3+x)/(3x^2-4)x=2 is
Solve for x: (x^3+x)/(3x^2-4)x=2
The solution is