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

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

Solve for x

$$x=-1$$

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

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{3} by 1-\frac{1}{2}x.

$$x^{2}+\frac{1}{2}x=\frac{1}{3}-\frac{1}{6}x$$

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

$$x^{2}+\frac{1}{2}x-\frac{1}{3}=-\frac{1}{6}x$$

Add \frac{1}{6}x to both sides.

$$x^{2}+\frac{1}{2}x-\frac{1}{3}+\frac{1}{6}x=0$$

Combine \frac{1}{2}x and \frac{1}{6}x to get \frac{2}{3}x.

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

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

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

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

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

Multiply -4 times -\frac{1}{3}.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+\frac{4}{3}}}{2}$$

Add \frac{4}{9} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{16}{9}}}{2}$$

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

$$x=\frac{-\frac{2}{3}±\frac{4}{3}}{2}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{4}{3}}{2} when ± is plus. Add -\frac{2}{3} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Now solve the equation x=\frac{-\frac{2}{3}±\frac{4}{3}}{2} when ± is minus. Subtract \frac{4}{3} from -\frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{3} by 1-\frac{1}{2}x.

$$x^{2}+\frac{1}{2}x=\frac{1}{3}-\frac{1}{6}x$$

Add \frac{1}{6}x to both sides.

$$x^{2}+\frac{1}{2}x+\frac{1}{6}x=\frac{1}{3}$$

Combine \frac{1}{2}x and \frac{1}{6}x to get \frac{2}{3}x.

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

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

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

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

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

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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