Solve (x-1)^2+(x-3)^2=25

$$( x - 1 ) ^ { 2 } + ( x - 3 ) ^ { 2 } = 25$$

Solve for x

$$x=\frac{\sqrt{46}}{2}+2\approx 5.391164992$$

$$x=-\frac{\sqrt{46}}{2}+2\approx -1.391164992$$

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}.

$$x^{2}-2x+1+\left(x-3\right)^{2}=25$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-2x+1+x^{2}-6x+9=25$$

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

$$2x^{2}-2x+1-6x+9=25$$

Combine -2x and -6x to get -8x.

$$2x^{2}-8x+1+9=25$$

Add 1 and 9 to get 10.

$$2x^{2}-8x+10=25$$

Subtract 25 from both sides.

$$2x^{2}-8x+10-25=0$$

Subtract 25 from 10 to get -15.

$$2x^{2}-8x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-\left(-8\right)±\sqrt{64+120}}{2\times 2}$$

Add 64 to 120.

$$x=\frac{-\left(-8\right)±\sqrt{184}}{2\times 2}$$

Take the square root of 184.

$$x=\frac{-\left(-8\right)±2\sqrt{46}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{46}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±2\sqrt{46}}{4}$$

Now solve the equation x=\frac{8±2\sqrt{46}}{4} when ± is plus. Add 8 to 2\sqrt{46}.

$$x=\frac{2\sqrt{46}+8}{4}$$

Divide 8+2\sqrt{46} by 4.

$$x=\frac{\sqrt{46}}{2}+2$$

Now solve the equation x=\frac{8±2\sqrt{46}}{4} when ± is minus. Subtract 2\sqrt{46} from 8.

$$x=\frac{8-2\sqrt{46}}{4}$$

Divide 8-2\sqrt{46} by 4.

$$x=-\frac{\sqrt{46}}{2}+2$$

The equation is now solved.

$$x=\frac{\sqrt{46}}{2}+2 x=-\frac{\sqrt{46}}{2}+2$$

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}.

$$x^{2}-2x+1+\left(x-3\right)^{2}=25$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-2x+1+x^{2}-6x+9=25$$

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

$$2x^{2}-2x+1-6x+9=25$$

Combine -2x and -6x to get -8x.

$$2x^{2}-8x+1+9=25$$

Add 1 and 9 to get 10.

$$2x^{2}-8x+10=25$$

Subtract 10 from both sides.

$$2x^{2}-8x=25-10$$

Subtract 10 from 25 to get 15.

$$2x^{2}-8x=15$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{15}{2}$$

Divide -8 by 2.

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

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.

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

Square -2.

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

Add \frac{15}{2} to 4.

$$x^{2}-4x+4=\frac{23}{2}$$

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}.

$$\left(x-2\right)^{2}=\frac{23}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{23}{2}}$$

Simplify.

$$x-2=\frac{\sqrt{46}}{2} x-2=-\frac{\sqrt{46}}{2}$$

Add 2 to both sides of the equation.

$$x=\frac{\sqrt{46}}{2}+2 x=-\frac{\sqrt{46}}{2}+2$$