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

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

Solve for x

$$x = \frac{\sqrt{13} + 5}{2} \approx 4.302775638$$

$$x=\frac{5-\sqrt{13}}{2}\approx 0.697224362$$

Steps Using the Quadratic Formula

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

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$x^{2}-4x+3=x$$

Subtract x from both sides.

$$x^{2}-4x+3-x=0$$

Combine -4x and -x to get -5x.

$$x^{2}-5x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12}}{2}$$

Add 25 to -12.

$$x=\frac{-\left(-5\right)±\sqrt{13}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{13}}{2}$$

Now solve the equation x=\frac{5±\sqrt{13}}{2} when ± is plus. Add 5 to \sqrt{13}.

$$x=\frac{\sqrt{13}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from 5.

$$x=\frac{5-\sqrt{13}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$

Steps for Completing the Square

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

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

Subtract x from both sides.

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

Combine -4x and -x to get -5x.

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

Subtract 4 from both sides.

$$x^{2}-5x=1-4$$

Subtract 4 from 1 to get -3.

$$x^{2}-5x=-3$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-3+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-3+\frac{25}{4}$$

Add -3 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{13}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{13}}{2} x-\frac{5}{2}=-\frac{\sqrt{13}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$