Solve 3/2x^2=-x^2+2/3x

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

Solve for x

$$x=\frac{4}{15}\approx 0.266666667$$

$$x=0$$

Steps Using Factoring

Add x^{2} to both sides.

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

Combine \frac{3}{2}x^{2} and x^{2} to get \frac{5}{2}x^{2}.

$$\frac{5}{2}x^{2}=\frac{2}{3}x$$

Subtract \frac{2}{3}x from both sides.

$$\frac{5}{2}x^{2}-\frac{2}{3}x=0$$

Factor out x.

$$x\left(\frac{5}{2}x-\frac{2}{3}\right)=0$$

To find equation solutions, solve x=0 and \frac{5x}{2}-\frac{2}{3}=0.

$$x=0 x=\frac{4}{15}$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

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

Combine \frac{3}{2}x^{2} and x^{2} to get \frac{5}{2}x^{2}.

$$\frac{5}{2}x^{2}=\frac{2}{3}x$$

Subtract \frac{2}{3}x from both sides.

$$\frac{5}{2}x^{2}-\frac{2}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{2} for a, -\frac{2}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\left(-\frac{2}{3}\right)^{2}}}{2\times \frac{5}{2}}$$

Take the square root of \left(-\frac{2}{3}\right)^{2}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\frac{2}{3}}{2\times \frac{5}{2}}$$

The opposite of -\frac{2}{3} is \frac{2}{3}.

$$x=\frac{\frac{2}{3}±\frac{2}{3}}{2\times \frac{5}{2}}$$

Multiply 2 times \frac{5}{2}.

$$x=\frac{\frac{2}{3}±\frac{2}{3}}{5}$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{2}{3}}{5} when ± is plus. Add \frac{2}{3} to \frac{2}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{4}{3}}{5}$$

Divide \frac{4}{3} by 5.

$$x=\frac{4}{15}$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{2}{3}}{5} when ± is minus. Subtract \frac{2}{3} from \frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{5}$$

Divide 0 by 5.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{15} x=0$$

Steps for Completing the Square

Add x^{2} to both sides.

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

Combine \frac{3}{2}x^{2} and x^{2} to get \frac{5}{2}x^{2}.

$$\frac{5}{2}x^{2}=\frac{2}{3}x$$

Subtract \frac{2}{3}x from both sides.

$$\frac{5}{2}x^{2}-\frac{2}{3}x=0$$

Divide both sides of the equation by \frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{2}x^{2}-\frac{2}{3}x}{\frac{5}{2}}=\frac{0}{\frac{5}{2}}$$

Dividing by \frac{5}{2} undoes the multiplication by \frac{5}{2}.

$$x^{2}+\left(-\frac{\frac{2}{3}}{\frac{5}{2}}\right)x=\frac{0}{\frac{5}{2}}$$

Divide -\frac{2}{3} by \frac{5}{2} by multiplying -\frac{2}{3} by the reciprocal of \frac{5}{2}.

$$x^{2}-\frac{4}{15}x=\frac{0}{\frac{5}{2}}$$

Divide 0 by \frac{5}{2} by multiplying 0 by the reciprocal of \frac{5}{2}.

$$x^{2}-\frac{4}{15}x=0$$

Divide -\frac{4}{15}, the coefficient of the x term, by 2 to get -\frac{2}{15}. Then add the square of -\frac{2}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{15}x+\left(-\frac{2}{15}\right)^{2}=\left(-\frac{2}{15}\right)^{2}$$

Square -\frac{2}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{15}x+\frac{4}{225}=\frac{4}{225}$$

Factor x^{2}-\frac{4}{15}x+\frac{4}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{15}\right)^{2}=\frac{4}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{15}\right)^{2}}=\sqrt{\frac{4}{225}}$$

Simplify.

$$x-\frac{2}{15}=\frac{2}{15} x-\frac{2}{15}=-\frac{2}{15}$$

Add \frac{2}{15} to both sides of the equation.

$$x=\frac{4}{15} x=0$$