Solve -4/3x=1/2x^2

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

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$x=0$$

Steps Using Factoring

Subtract \frac{1}{2}x^{2} from both sides.

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

Factor out x.

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

To find equation solutions, solve x=0 and -\frac{4}{3}-\frac{x}{2}=0.

$$x=0 x=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x^{2} from both sides.

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

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.

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

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

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

Take the square root of \left(-\frac{4}{3}\right)^{2}.

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

The opposite of -\frac{4}{3} is \frac{4}{3}.

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

Multiply 2 times -\frac{1}{2}.

$$x=\frac{\frac{4}{3}±\frac{4}{3}}{-1}$$

Now solve the equation x=\frac{\frac{4}{3}±\frac{4}{3}}{-1} when ± is plus. Add \frac{4}{3} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{8}{3}}{-1}$$

Divide \frac{8}{3} by -1.

$$x=-\frac{8}{3}$$

Now solve the equation x=\frac{\frac{4}{3}±\frac{4}{3}}{-1} when ± is minus. Subtract \frac{4}{3} from \frac{4}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{-1}$$

Divide 0 by -1.

$$x=0$$

The equation is now solved.

$$x=-\frac{8}{3} x=0$$

Steps for Completing the Square

Subtract \frac{1}{2}x^{2} from both sides.

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

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.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide -\frac{4}{3} by -\frac{1}{2} by multiplying -\frac{4}{3} by the reciprocal of -\frac{1}{2}.

$$x^{2}+\frac{8}{3}x=\frac{0}{-\frac{1}{2}}$$

Divide 0 by -\frac{1}{2} by multiplying 0 by the reciprocal of -\frac{1}{2}.

$$x^{2}+\frac{8}{3}x=0$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{16}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{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}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{4}{3}=\frac{4}{3} x+\frac{4}{3}=-\frac{4}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

$$x=0 x=-\frac{8}{3}$$