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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{2}i}{2}+1\approx 1+0.707106781i$$

$$x=-\frac{\sqrt{2}i}{2}+1\approx 1-0.707106781i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

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

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

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

Subtract 2 from both sides.

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

Subtract 2 from -1 to get -3.

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-4±\sqrt{16-24}}{2\left(-2\right)}$$

Add 16 to -24.

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

Take the square root of -8.

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

Multiply 2 times -2.

$$x=\frac{-4±2\sqrt{2}i}{-4}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

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

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

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

Add 1 to both sides.

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

Add 2 and 1 to get 3.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

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

Divide 3 by -2.

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

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

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

Add -\frac{3}{2} to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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