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

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

Solve for w

$$w=1$$

$$w=0$$

Steps Using Factoring

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

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

Combine -2w and w to get -w.

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

Subtract 1 from both sides.

$$1-w+w^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$-w+w^{2}=0$$

Factor out w.

$$w\left(-1+w\right)=0$$

To find equation solutions, solve w=0 and -1+w=0.

$$w=0 w=1$$

Steps Using the Quadratic Formula

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

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

Combine -2w and w to get -w.

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

Subtract 1 from both sides.

$$1-w+w^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$-w+w^{2}=0$$

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.

$$w^{2}-w=0$$

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

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

Take the square root of 1.

$$w=\frac{-\left(-1\right)±1}{2}$$

The opposite of -1 is 1.

$$w=\frac{1±1}{2}$$

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

$$w=\frac{2}{2}$$

Divide 2 by 2.

$$w=1$$

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

$$w=\frac{0}{2}$$

Divide 0 by 2.

$$w=0$$

The equation is now solved.

$$w=1 w=0$$

Steps for Completing the Square

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

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

Combine -2w and w to get -w.

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

Subtract 1 from both sides.

$$1-w+w^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$-w+w^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$w^{2}-w=0$$

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.

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

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

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

Factor w^{2}-w+\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}.

$$\left(w-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$w-\frac{1}{2}=\frac{1}{2} w-\frac{1}{2}=-\frac{1}{2}$$

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

$$w=1 w=0$$