Solve (1/3x-2)^2=(x-2/3)^2

$$( \frac { 1 } { 3 } x - 2 ) ^ { 2 } = ( x - \frac { 2 } { 3 } ) ^ { 2 }$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{3}x-2\right)^{2}.

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4=\left(x-\frac{2}{3}\right)^{2}$$

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

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4=x^{2}-\frac{4}{3}x+\frac{4}{9}$$

Subtract x^{2} from both sides.

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4-x^{2}=-\frac{4}{3}x+\frac{4}{9}$$

Combine \frac{1}{9}x^{2} and -x^{2} to get -\frac{8}{9}x^{2}.

$$-\frac{8}{9}x^{2}-\frac{4}{3}x+4=-\frac{4}{3}x+\frac{4}{9}$$

Add \frac{4}{3}x to both sides.

$$-\frac{8}{9}x^{2}-\frac{4}{3}x+4+\frac{4}{3}x=\frac{4}{9}$$

Combine -\frac{4}{3}x and \frac{4}{3}x to get 0.

$$-\frac{8}{9}x^{2}+4=\frac{4}{9}$$

Subtract 4 from both sides.

$$-\frac{8}{9}x^{2}=\frac{4}{9}-4$$

Subtract 4 from \frac{4}{9} to get -\frac{32}{9}.

$$-\frac{8}{9}x^{2}=-\frac{32}{9}$$

Multiply both sides by -\frac{9}{8}, the reciprocal of -\frac{8}{9}.

$$x^{2}=-\frac{32}{9}\left(-\frac{9}{8}\right)$$

Multiply -\frac{32}{9} and -\frac{9}{8} to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{3}x-2\right)^{2}.

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4=\left(x-\frac{2}{3}\right)^{2}$$

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

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4=x^{2}-\frac{4}{3}x+\frac{4}{9}$$

Subtract x^{2} from both sides.

$$\frac{1}{9}x^{2}-\frac{4}{3}x+4-x^{2}=-\frac{4}{3}x+\frac{4}{9}$$

Combine \frac{1}{9}x^{2} and -x^{2} to get -\frac{8}{9}x^{2}.

$$-\frac{8}{9}x^{2}-\frac{4}{3}x+4=-\frac{4}{3}x+\frac{4}{9}$$

Add \frac{4}{3}x to both sides.

$$-\frac{8}{9}x^{2}-\frac{4}{3}x+4+\frac{4}{3}x=\frac{4}{9}$$

Combine -\frac{4}{3}x and \frac{4}{3}x to get 0.

$$-\frac{8}{9}x^{2}+4=\frac{4}{9}$$

Subtract \frac{4}{9} from both sides.

$$-\frac{8}{9}x^{2}+4-\frac{4}{9}=0$$

Subtract \frac{4}{9} from 4 to get \frac{32}{9}.

$$-\frac{8}{9}x^{2}+\frac{32}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{8}{9} for a, 0 for b, and \frac{32}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{8}{9}\right)\times \frac{32}{9}}}{2\left(-\frac{8}{9}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{8}{9}\right)\times \frac{32}{9}}}{2\left(-\frac{8}{9}\right)}$$

Multiply -4 times -\frac{8}{9}.

$$x=\frac{0±\sqrt{\frac{32}{9}\times \frac{32}{9}}}{2\left(-\frac{8}{9}\right)}$$

Multiply \frac{32}{9} times \frac{32}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0±\sqrt{\frac{1024}{81}}}{2\left(-\frac{8}{9}\right)}$$

Take the square root of \frac{1024}{81}.

$$x=\frac{0±\frac{32}{9}}{2\left(-\frac{8}{9}\right)}$$

Multiply 2 times -\frac{8}{9}.

$$x=\frac{0±\frac{32}{9}}{-\frac{16}{9}}$$

Now solve the equation x=\frac{0±\frac{32}{9}}{-\frac{16}{9}} when ± is plus.

$$x=-2$$

Now solve the equation x=\frac{0±\frac{32}{9}}{-\frac{16}{9}} when ± is minus.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$