Solve (x-3)^2=x-2

$$( x - 3 ) ^ { 2 } = x - 2$$

Solve for x

$$x = \frac{\sqrt{5} + 7}{2} \approx 4.618033989$$

$$x = \frac{7 - \sqrt{5}}{2} \approx 2.381966011$$

Steps Using the Quadratic Formula

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

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

Subtract x from both sides.

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

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-2$$

Add 2 to both sides.

$$x^{2}-7x+9+2=0$$

Add 9 and 2 to get 11.

$$x^{2}-7x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 11}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 11}}{2}$$

Multiply -4 times 11.

$$x=\frac{-\left(-7\right)±\sqrt{49-44}}{2}$$

Add 49 to -44.

$$x=\frac{-\left(-7\right)±\sqrt{5}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{5}}{2}$$

Now solve the equation x=\frac{7±\sqrt{5}}{2} when ± is plus. Add 7 to \sqrt{5}.

$$x=\frac{\sqrt{5}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 7.

$$x=\frac{7-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+7}{2} x=\frac{7-\sqrt{5}}{2}$$

Steps for Completing the Square

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

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

Subtract x from both sides.

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

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-2$$

Subtract 9 from both sides.

$$x^{2}-7x=-2-9$$

Subtract 9 from -2 to get -11.

$$x^{2}-7x=-11$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-11+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-11+\frac{49}{4}$$

Add -11 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{5}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{5}}{2} x-\frac{7}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+7}{2} x=\frac{7-\sqrt{5}}{2}$$