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

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

Solve for x

$$x=\sqrt{2}+2\approx 3.414213562$$

$$x=2-\sqrt{2}\approx 0.585786438$$

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}.

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

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

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

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

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

Square 4.

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

Multiply -4 times -1.

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

Multiply 4 times -2.

$$x=\frac{-4±\sqrt{16-8}}{2\left(-1\right)}$$

Add 16 to -8.

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

Take the square root of 8.

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

Multiply 2 times -1.

$$x=\frac{-4±2\sqrt{2}}{-2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

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}.

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

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

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 4 by -1.

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

Divide 2 by -1.

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

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.

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

Square -2.

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

Add -2 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

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