Solve (x+1/3)(x-1/3)=1/3

$$( x + 1 / 3 ) ( x - 1 / 3 ) = 1 / 3$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps by Finding Square Root

Consider \left(x+\frac{1}{3}\right)\left(x-\frac{1}{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{1}{3}.

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

Add \frac{1}{9} to both sides.

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

Add \frac{1}{3} and \frac{1}{9} to get \frac{4}{9}.

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

Take the square root of both sides of the equation.

$$x=\frac{2}{3} x=-\frac{2}{3}$$

Steps Using the Quadratic Formula

Consider \left(x+\frac{1}{3}\right)\left(x-\frac{1}{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{1}{3}.

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

Subtract \frac{1}{3} from both sides.

$$x^{2}-\frac{1}{9}-\frac{1}{3}=0$$

Subtract \frac{1}{3} from -\frac{1}{9} to get -\frac{4}{9}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{4}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 0.

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

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

$$x=\frac{0±\sqrt{\frac{16}{9}}}{2}$$

Take the square root of \frac{16}{9}.

$$x=\frac{0±\frac{4}{3}}{2}$$

Now solve the equation x=\frac{0±\frac{4}{3}}{2} when ± is plus.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{0±\frac{4}{3}}{2} when ± is minus.

$$x=-\frac{2}{3}$$

The equation is now solved.

$$x=\frac{2}{3} x=-\frac{2}{3}$$