Solve x=(x^2+4)*1/2x+2

$$x = ( x ^ { 2 } + 4 ) \cdot \frac { 1 } { 2 x } + 2$$

Solve for x

$$x=2\sqrt{2}+2\approx 4.828427125$$

$$x=2-2\sqrt{2}\approx -0.828427125$$

Steps Using the Quadratic Formula

Express \left(x^{2}+4\right)\times \frac{1}{2x} as a single fraction.

$$x=\frac{x^{2}+4}{2x}+2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2x}{2x}.

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

Since \frac{x^{2}+4}{2x} and \frac{2\times 2x}{2x} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x^{2}+4+2\times 2x.

$$x=\frac{x^{2}+4+4x}{2x}$$

Subtract \frac{x^{2}+4+4x}{2x} from both sides.

$$x-\frac{x^{2}+4+4x}{2x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2x}{2x}.

$$\frac{x\times 2x}{2x}-\frac{x^{2}+4+4x}{2x}=0$$

Since \frac{x\times 2x}{2x} and \frac{x^{2}+4+4x}{2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 2x-\left(x^{2}+4+4x\right)}{2x}=0$$

Do the multiplications in x\times 2x-\left(x^{2}+4+4x\right).

$$\frac{2x^{2}-x^{2}-4-4x}{2x}=0$$

Combine like terms in 2x^{2}-x^{2}-4-4x.

$$\frac{x^{2}-4-4x}{2x}=0$$

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

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+16}}{2}$$

Add 16 to 16.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2}$$

Take the square root of 32.

$$x=\frac{-\left(-4\right)±4\sqrt{2}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{4±4\sqrt{2}}{2} when ± is plus. Add 4 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+4}{2}$$

Divide 4+4\sqrt{2} by 2.

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

Now solve the equation x=\frac{4±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from 4.

$$x=\frac{4-4\sqrt{2}}{2}$$

Divide 4-4\sqrt{2} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Express \left(x^{2}+4\right)\times \frac{1}{2x} as a single fraction.

$$x=\frac{x^{2}+4}{2x}+2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2x}{2x}.

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

Since \frac{x^{2}+4}{2x} and \frac{2\times 2x}{2x} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x^{2}+4+2\times 2x.

$$x=\frac{x^{2}+4+4x}{2x}$$

Subtract \frac{x^{2}+4+4x}{2x} from both sides.

$$x-\frac{x^{2}+4+4x}{2x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2x}{2x}.

$$\frac{x\times 2x}{2x}-\frac{x^{2}+4+4x}{2x}=0$$

Since \frac{x\times 2x}{2x} and \frac{x^{2}+4+4x}{2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 2x-\left(x^{2}+4+4x\right)}{2x}=0$$

Do the multiplications in x\times 2x-\left(x^{2}+4+4x\right).

$$\frac{2x^{2}-x^{2}-4-4x}{2x}=0$$

Combine like terms in 2x^{2}-x^{2}-4-4x.

$$\frac{x^{2}-4-4x}{2x}=0$$

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

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

Add 4 to both sides. Anything plus zero gives itself.

$$x^{2}-4x=4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=4+\left(-2\right)^{2}$$

Square -2.

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

Add 4 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{8}$$

Simplify.

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

Add 2 to both sides of the equation.

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