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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{2}i}{4}+\frac{1}{2}\approx 0.5+0.353553391i$$

$$x=-\frac{\sqrt{2}i}{4}+\frac{1}{2}\approx 0.5-0.353553391i$$

Steps Using the Quadratic Formula

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Subtract \frac{1}{4} from both sides.

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

Subtract \frac{1}{4} from 1 to get \frac{3}{4}.

$$2x^{2}-2x+\frac{3}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and \frac{3}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\times \frac{3}{4}}}{2\times 2}$$

Multiply -8 times \frac{3}{4}.

$$x=\frac{-\left(-2\right)±\sqrt{4-6}}{2\times 2}$$

Add 4 to -6.

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

Take the square root of -2.

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

The opposite of -2 is 2.

$$x=\frac{2±\sqrt{2}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±\sqrt{2}i}{4}$$

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

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

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

$$x=\frac{\sqrt{2}i}{4}+\frac{1}{2}$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from \frac{1}{4} to get -\frac{3}{4}.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide -\frac{3}{4} by 2.

$$x^{2}-x=-\frac{3}{8}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{3}{8}+\frac{1}{4}$$

Add -\frac{3}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{8}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{1}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{1}{8}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

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