Solve 1/2+1/2x^2=2x

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

Solve for x

$$x=\sqrt{3}+2\approx 3.732050808$$

$$x=2-\sqrt{3}\approx 0.267949192$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

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.

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

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

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

Square -2.

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

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

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

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

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

Add 4 to -1.

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

The opposite of -2 is 2.

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

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

$$x=\frac{2±\sqrt{3}}{1}$$

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

$$x=\frac{\sqrt{3}+2}{1}$$

Divide 2+\sqrt{3} by 1.

$$x=\sqrt{3}+2$$

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

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

Divide 2-\sqrt{3} by 1.

$$x=2-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}+2 x=2-\sqrt{3}$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 2.

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

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

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

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

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

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

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

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.

$$x^{2}-4x+\left(-2\right)^{2}=-1+\left(-2\right)^{2}$$

Square -2.

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

Add -1 to 4.

$$x^{2}-4x+4=3$$

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}.

$$\left(x-2\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{3}$$

Simplify.

$$x-2=\sqrt{3} x-2=-\sqrt{3}$$

Add 2 to both sides of the equation.

$$x=\sqrt{3}+2 x=2-\sqrt{3}$$