Solve (x-3/2)^2=1/3-2/9

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

Solve for x

$$x = \frac{11}{6} = 1\frac{5}{6} \approx 1.833333333$$

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

Steps Using the Quadratic Formula

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

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

Subtract \frac{2}{9} from \frac{1}{3} to get \frac{1}{9}.

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

Subtract \frac{1}{9} from both sides.

$$x^{2}-3x+\frac{9}{4}-\frac{1}{9}=0$$

Subtract \frac{1}{9} from \frac{9}{4} to get \frac{77}{36}.

$$x^{2}-3x+\frac{77}{36}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and \frac{77}{36} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{77}{36}}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{77}{36}}}{2}$$

Multiply -4 times \frac{77}{36}.

$$x=\frac{-\left(-3\right)±\sqrt{9-\frac{77}{9}}}{2}$$

Add 9 to -\frac{77}{9}.

$$x=\frac{-\left(-3\right)±\sqrt{\frac{4}{9}}}{2}$$

Take the square root of \frac{4}{9}.

$$x=\frac{-\left(-3\right)±\frac{2}{3}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\frac{2}{3}}{2}$$

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

$$x=\frac{\frac{11}{3}}{2}$$

Divide \frac{11}{3} by 2.

$$x=\frac{11}{6}$$

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

$$x=\frac{\frac{7}{3}}{2}$$

Divide \frac{7}{3} by 2.

$$x=\frac{7}{6}$$

The equation is now solved.

$$x=\frac{11}{6} x=\frac{7}{6}$$

Steps for Completing the Square

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

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

Subtract \frac{2}{9} from \frac{1}{3} to get \frac{1}{9}.

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

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{11}{6} x=\frac{7}{6}$$