Solve (x-8)^2=x

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

Solve for x

$$x = \frac{\sqrt{33} + 17}{2} \approx 11.372281323$$

$$x = \frac{17 - \sqrt{33}}{2} \approx 5.627718677$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-8\right)^{2}.

$$x^{2}-16x+64=x$$

Subtract x from both sides.

$$x^{2}-16x+64-x=0$$

Combine -16x and -x to get -17x.

$$x^{2}-17x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -17 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 64}}{2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 64}}{2}$$

Multiply -4 times 64.

$$x=\frac{-\left(-17\right)±\sqrt{289-256}}{2}$$

Add 289 to -256.

$$x=\frac{-\left(-17\right)±\sqrt{33}}{2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{33}}{2}$$

Now solve the equation x=\frac{17±\sqrt{33}}{2} when ± is plus. Add 17 to \sqrt{33}.

$$x=\frac{\sqrt{33}+17}{2}$$

Now solve the equation x=\frac{17±\sqrt{33}}{2} when ± is minus. Subtract \sqrt{33} from 17.

$$x=\frac{17-\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+17}{2} x=\frac{17-\sqrt{33}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-8\right)^{2}.

$$x^{2}-16x+64=x$$

Subtract x from both sides.

$$x^{2}-16x+64-x=0$$

Combine -16x and -x to get -17x.

$$x^{2}-17x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-17x=-64$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=-64+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=-64+\frac{289}{4}$$

Add -64 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{33}{4}$$

Factor x^{2}-17x+\frac{289}{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}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{\sqrt{33}}{2} x-\frac{17}{2}=-\frac{\sqrt{33}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{\sqrt{33}+17}{2} x=\frac{17-\sqrt{33}}{2}$$