Solve sqrt{2x}+2=1/2x

$$\sqrt { 2 x } + 2 = \frac { 1 } { 2 } x$$

Solve for x

$$x=4\left(\sqrt{3}+2\right)\approx 14.92820323$$

Solution Steps

Subtract 2 from both sides of the equation.

$$\sqrt{2x}=\frac{1}{2}x-2$$

Square both sides of the equation.

$$\left(\sqrt{2x}\right)^{2}=\left(\frac{1}{2}x-2\right)^{2}$$

Calculate \sqrt{2x} to the power of 2 and get 2x.

$$2x=\left(\frac{1}{2}x-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-2\right)^{2}.

$$2x=\frac{1}{4}x^{2}-2x+4$$

Subtract \frac{1}{4}x^{2} from both sides.

$$2x-\frac{1}{4}x^{2}=-2x+4$$

Add 2x to both sides.

$$2x-\frac{1}{4}x^{2}+2x=4$$

Combine 2x and 2x to get 4x.

$$4x-\frac{1}{4}x^{2}=4$$

Subtract 4 from both sides.

$$4x-\frac{1}{4}x^{2}-4=0$$

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.

$$-\frac{1}{4}x^{2}+4x-4=0$$

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

$$x=\frac{-4±\sqrt{4^{2}-4\left(-\frac{1}{4}\right)\left(-4\right)}}{2\left(-\frac{1}{4}\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{1}{4}\right)\left(-4\right)}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$x=\frac{-4±\sqrt{16-4}}{2\left(-\frac{1}{4}\right)}$$

Add 16 to -4.

$$x=\frac{-4±\sqrt{12}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of 12.

$$x=\frac{-4±2\sqrt{3}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} when ± is plus. Add -4 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-4}{-\frac{1}{2}}$$

Divide -4+2\sqrt{3} by -\frac{1}{2} by multiplying -4+2\sqrt{3} by the reciprocal of -\frac{1}{2}.

$$x=8-4\sqrt{3}$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} when ± is minus. Subtract 2\sqrt{3} from -4.

$$x=\frac{-2\sqrt{3}-4}{-\frac{1}{2}}$$

Divide -4-2\sqrt{3} by -\frac{1}{2} by multiplying -4-2\sqrt{3} by the reciprocal of -\frac{1}{2}.

$$x=4\sqrt{3}+8$$

The equation is now solved.

$$x=8-4\sqrt{3} x=4\sqrt{3}+8$$

Substitute 8-4\sqrt{3} for x in the equation \sqrt{2x}+2=\frac{1}{2}x.

$$\sqrt{2\left(8-4\sqrt{3}\right)}+2=\frac{1}{2}\left(8-4\sqrt{3}\right)$$

Simplify. The value x=8-4\sqrt{3} does not satisfy the equation.

$$2\times 3^{\frac{1}{2}}=4-2\times 3^{\frac{1}{2}}$$

Substitute 4\sqrt{3}+8 for x in the equation \sqrt{2x}+2=\frac{1}{2}x.

$$\sqrt{2\left(4\sqrt{3}+8\right)}+2=\frac{1}{2}\left(4\sqrt{3}+8\right)$$

Simplify. The value x=4\sqrt{3}+8 satisfies the equation.

$$2\times 3^{\frac{1}{2}}+4=2\times 3^{\frac{1}{2}}+4$$

Equation \sqrt{2x}=\frac{x}{2}-2 has a unique solution.

$$x=4\sqrt{3}+8$$