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

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

Solve for x (complex solution)

$$x=\frac{5+\sqrt{19}i}{2}\approx 2.5+2.179449472i$$

$$x=\frac{-\sqrt{19}i+5}{2}\approx 2.5-2.179449472i$$

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}.

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

Subtract 8 from 1 to get -7.

$$x^{2}-4x+4=-7+x$$

Subtract -7 from both sides.

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

The opposite of -7 is 7.

$$x^{2}-4x+4+7=x$$

Subtract x from both sides.

$$x^{2}-4x+4+7-x=0$$

Add 4 and 7 to get 11.

$$x^{2}-4x+11-x=0$$

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

$$x^{2}-5x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 11}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 11}}{2}$$

Multiply -4 times 11.

$$x=\frac{-\left(-5\right)±\sqrt{25-44}}{2}$$

Add 25 to -44.

$$x=\frac{-\left(-5\right)±\sqrt{-19}}{2}$$

Take the square root of -19.

$$x=\frac{-\left(-5\right)±\sqrt{19}i}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{19}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{19}i}{2} when ± is plus. Add 5 to i\sqrt{19}.

$$x=\frac{5+\sqrt{19}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{19}i}{2} when ± is minus. Subtract i\sqrt{19} from 5.

$$x=\frac{-\sqrt{19}i+5}{2}$$

The equation is now solved.

$$x=\frac{5+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+5}{2}$$

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}.

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

Subtract 8 from 1 to get -7.

$$x^{2}-4x+4=-7+x$$

Subtract x from both sides.

$$x^{2}-4x+4-x=-7$$

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

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

Subtract 4 from both sides.

$$x^{2}-5x=-7-4$$

Subtract 4 from -7 to get -11.

$$x^{2}-5x=-11$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-11+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-11+\frac{25}{4}$$

Add -11 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=-\frac{19}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{19}i}{2} x-\frac{5}{2}=-\frac{\sqrt{19}i}{2}$$

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

$$x=\frac{5+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+5}{2}$$