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

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Factor out x.

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

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

$$x=0 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

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

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

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

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

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

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

Multiply 2 times -2.

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

Now solve the equation x=\frac{\frac{2}{3}±\frac{2}{3}}{-4} 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.

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

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

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

Now solve the equation x=\frac{\frac{2}{3}±\frac{2}{3}}{-4} 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.

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

Divide 0 by -4.

$$x=0$$

The equation is now solved.

$$x=-\frac{1}{3} x=0$$

Steps for Completing the Square

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

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

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

Divide 0 by -2.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{6} from both sides of the equation.

$$x=0 x=-\frac{1}{3}$$