Solve 9*{left(x+1/3right)}^2=4

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

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-1$$

Steps by Finding Square Root

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

Take the square root of both sides of the equation.

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

Subtract \frac{1}{3} from both sides of the equation.

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

Subtracting \frac{1}{3} from itself leaves 0.

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

Subtract \frac{1}{3} from \frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1}{3}$$

Subtract \frac{1}{3} from -\frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-1$$

The equation is now solved.

$$x=\frac{1}{3} x=-1$$