Solve x^2+left(-x+1right)^2=3

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

Solve for x

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

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}.

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

Calculate -x to the power of 2 and get x^{2}.

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

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

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

Subtract 3 from both sides.

$$2x^{2}+2\left(-x\right)+1-3=0$$

Subtract 3 from 1 to get -2.

$$2x^{2}+2\left(-x\right)-2=0$$

Multiply 2 and -1 to get -2.

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

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

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

Square -2.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

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

Add 4 to 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\times 2}$$

Take the square root of 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±2\sqrt{5}}{4}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{4} when ± is plus. Add 2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{4}$$

Divide 2+2\sqrt{5} by 4.

$$x=\frac{\sqrt{5}+1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{4}$$

Divide 2-2\sqrt{5} by 4.

$$x=\frac{1-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

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}.

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

Calculate -x to the power of 2 and get x^{2}.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 3 to get 2.

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

Multiply 2 and -1 to get -2.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide 2 by 2.

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

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.

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

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

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

Add 1 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$