Solve (x-1)^2+(4-3)^2=2

$$( x - 1 ) ^ { 2 } + ( 4 - 3 ) ^ { 2 } = 2$$

Solve for x

$$x=2$$

$$x=0$$

Steps Using Factoring

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

$$x^{2}-2x+1+\left(4-3\right)^{2}=2$$

Subtract 3 from 4 to get 1.

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

Calculate 1 to the power of 2 and get 1.

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

Add 1 and 1 to get 2.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

Factor out x.

$$x\left(x-2\right)=0$$

To find equation solutions, solve x=0 and x-2=0.

$$x=0 x=2$$

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}.

$$x^{2}-2x+1+\left(4-3\right)^{2}=2$$

Subtract 3 from 4 to get 1.

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

Calculate 1 to the power of 2 and get 1.

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

Add 1 and 1 to get 2.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

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

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

Take the square root of \left(-2\right)^{2}.

$$x=\frac{-\left(-2\right)±2}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2}{2}$$

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

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

Divide 4 by 2.

$$x=2$$

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

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=2 x=0$$

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}.

$$x^{2}-2x+1+\left(4-3\right)^{2}=2$$

Subtract 3 from 4 to get 1.

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

Calculate 1 to the power of 2 and get 1.

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

Add 1 and 1 to get 2.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=1 x-1=-1$$

Add 1 to both sides of the equation.

$$x=2 x=0$$