Solve (x-1)^2+(x-2)^2=8

$$( x - 1 ) ^ { 2 } + ( x - 2 ) ^ { 2 } = 8$$

Solve for x

$$x = \frac{\sqrt{15} + 3}{2} \approx 3.436491673$$

$$x=\frac{3-\sqrt{15}}{2}\approx -0.436491673$$

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}.

$$x^{2}-2x+1+\left(x-2\right)^{2}=8$$

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

$$x^{2}-2x+1+x^{2}-4x+4=8$$

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

$$2x^{2}-2x+1-4x+4=8$$

Combine -2x and -4x to get -6x.

$$2x^{2}-6x+1+4=8$$

Add 1 and 4 to get 5.

$$2x^{2}-6x+5=8$$

Subtract 8 from both sides.

$$2x^{2}-6x+5-8=0$$

Subtract 8 from 5 to get -3.

$$2x^{2}-6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+24}}{2\times 2}$$

Add 36 to 24.

$$x=\frac{-\left(-6\right)±\sqrt{60}}{2\times 2}$$

Take the square root of 60.

$$x=\frac{-\left(-6\right)±2\sqrt{15}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{15}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2\sqrt{15}}{4}$$

Now solve the equation x=\frac{6±2\sqrt{15}}{4} when ± is plus. Add 6 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}+6}{4}$$

Divide 6+2\sqrt{15} by 4.

$$x=\frac{\sqrt{15}+3}{2}$$

Now solve the equation x=\frac{6±2\sqrt{15}}{4} when ± is minus. Subtract 2\sqrt{15} from 6.

$$x=\frac{6-2\sqrt{15}}{4}$$

Divide 6-2\sqrt{15} by 4.

$$x=\frac{3-\sqrt{15}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}+3}{2} x=\frac{3-\sqrt{15}}{2}$$

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}.

$$x^{2}-2x+1+\left(x-2\right)^{2}=8$$

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

$$x^{2}-2x+1+x^{2}-4x+4=8$$

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

$$2x^{2}-2x+1-4x+4=8$$

Combine -2x and -4x to get -6x.

$$2x^{2}-6x+1+4=8$$

Add 1 and 4 to get 5.

$$2x^{2}-6x+5=8$$

Subtract 5 from both sides.

$$2x^{2}-6x=8-5$$

Subtract 5 from 8 to get 3.

$$2x^{2}-6x=3$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{3}{2}$$

Divide -6 by 2.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{3}{2}+\frac{9}{4}$$

Add \frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{15}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{15}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{15}}{2} x-\frac{3}{2}=-\frac{\sqrt{15}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{15}+3}{2} x=\frac{3-\sqrt{15}}{2}$$