Solve 2x^2+4x(2/x^2)=8

$$2 { x }^{ 2 } +4x( \frac{ 2 }{ { x }^{ 2 } } )=8$$

Solve for x

$$x = -\frac{\sqrt[3]{2} \cdot 9 ^ {\frac{2}{3}} {(\sqrt[3]{\sqrt{33} + 9} + \sqrt[3]{9 - \sqrt{33}})}}{9} \approx -2.382975768$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$2x^{2}x^{2}+4x\times 2=8x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2x^{4}+4x\times 2=8x^{2}$$

Multiply 4 and 2 to get 8.

$$2x^{4}+8x=8x^{2}$$

Subtract 8x^{2} from both sides.

$$2x^{4}+8x-8x^{2}=0$$

Substitute t for x^{2}.

$$2t^{2}-8t+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -8 for b, and 8 for c in the quadratic formula.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 8}}{2\times 2}$$

Do the calculations.

$$t=\frac{8±0}{4}$$

Solutions are the same.

$$t=2$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=-\sqrt{2} x=\sqrt{2}$$