Solve -x(1/3-x)=4/9

$$- x ( \frac { 1 } { 3 } - x ) = \frac { 4 } { 9 }$$

Solve for x

$$x=\frac{\sqrt{17}+1}{6}\approx 0.853850938$$

$$x=\frac{1-\sqrt{17}}{6}\approx -0.520517604$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -x by \frac{1}{3}-x.

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

Multiply -1 and -1 to get 1.

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

Multiply x and x to get x^{2}.

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

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

$$\left(-x\right)\times \frac{1}{3}+x^{2}-\frac{4}{9}=0$$

Multiply -1 and \frac{1}{3} to get -\frac{1}{3}.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{1}{3} for b, and -\frac{4}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

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

$$x=\frac{-\left(-\frac{1}{3}\right)±\sqrt{\frac{1+16}{9}}}{2}$$

Add \frac{1}{9} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{3}\right)±\sqrt{\frac{17}{9}}}{2}$$

Take the square root of \frac{17}{9}.

$$x=\frac{-\left(-\frac{1}{3}\right)±\frac{\sqrt{17}}{3}}{2}$$

The opposite of -\frac{1}{3} is \frac{1}{3}.

$$x=\frac{\frac{1}{3}±\frac{\sqrt{17}}{3}}{2}$$

Now solve the equation x=\frac{\frac{1}{3}±\frac{\sqrt{17}}{3}}{2} when ± is plus. Add \frac{1}{3} to \frac{\sqrt{17}}{3}.

$$x=\frac{\sqrt{17}+1}{2\times 3}$$

Divide \frac{1+\sqrt{17}}{3} by 2.

$$x=\frac{\sqrt{17}+1}{6}$$

Now solve the equation x=\frac{\frac{1}{3}±\frac{\sqrt{17}}{3}}{2} when ± is minus. Subtract \frac{\sqrt{17}}{3} from \frac{1}{3}.

$$x=\frac{1-\sqrt{17}}{2\times 3}$$

Divide \frac{1-\sqrt{17}}{3} by 2.

$$x=\frac{1-\sqrt{17}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+1}{6} x=\frac{1-\sqrt{17}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply -x by \frac{1}{3}-x.

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

Multiply -1 and -1 to get 1.

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

Multiply x and x to get x^{2}.

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

Multiply -1 and \frac{1}{3} to get -\frac{1}{3}.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{4}{9}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{4}{9}+\frac{1}{36}$$

Add \frac{4}{9} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{17}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{17}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{17}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{17}}{6} x-\frac{1}{6}=-\frac{\sqrt{17}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{17}+1}{6} x=\frac{1-\sqrt{17}}{6}$$