Solve 4(1/3)+6x_{3}+(1/3)*x_3^2=-6

$$4 ( \frac { 1 } { 3 } ) + 6 x _ { 3 } + ( \frac { 1 } { 3 } ) \cdot x _ { 3 } ^ { 2 } = - 6$$

Solve for x_3

$$x_{3}=\sqrt{59}-9\approx -1.318854252$$

$$x_{3}=-\sqrt{59}-9\approx -16.681145748$$

Steps Using the Quadratic Formula

Multiply 4 and \frac{1}{3} to get \frac{4}{3}.

$$\frac{4}{3}+6x_{3}+\frac{1}{3}x_{3}^{2}=-6$$

Add 6 to both sides.

$$\frac{4}{3}+6x_{3}+\frac{1}{3}x_{3}^{2}+6=0$$

Add \frac{4}{3} and 6 to get \frac{22}{3}.

$$\frac{22}{3}+6x_{3}+\frac{1}{3}x_{3}^{2}=0$$

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.

$$\frac{1}{3}x_{3}^{2}+6x_{3}+\frac{22}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, 6 for b, and \frac{22}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x_{3}=\frac{-6±\sqrt{6^{2}-4\times \frac{1}{3}\times \frac{22}{3}}}{2\times \frac{1}{3}}$$

Square 6.

$$x_{3}=\frac{-6±\sqrt{36-4\times \frac{1}{3}\times \frac{22}{3}}}{2\times \frac{1}{3}}$$

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

$$x_{3}=\frac{-6±\sqrt{36-\frac{4}{3}\times \frac{22}{3}}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times \frac{22}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x_{3}=\frac{-6±\sqrt{36-\frac{88}{9}}}{2\times \frac{1}{3}}$$

Add 36 to -\frac{88}{9}.

$$x_{3}=\frac{-6±\sqrt{\frac{236}{9}}}{2\times \frac{1}{3}}$$

Take the square root of \frac{236}{9}.

$$x_{3}=\frac{-6±\frac{2\sqrt{59}}{3}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x_{3}=\frac{-6±\frac{2\sqrt{59}}{3}}{\frac{2}{3}}$$

Now solve the equation x_{3}=\frac{-6±\frac{2\sqrt{59}}{3}}{\frac{2}{3}} when ± is plus. Add -6 to \frac{2\sqrt{59}}{3}.

$$x_{3}=\frac{\frac{2\sqrt{59}}{3}-6}{\frac{2}{3}}$$

Divide -6+\frac{2\sqrt{59}}{3} by \frac{2}{3} by multiplying -6+\frac{2\sqrt{59}}{3} by the reciprocal of \frac{2}{3}.

$$x_{3}=\sqrt{59}-9$$

Now solve the equation x_{3}=\frac{-6±\frac{2\sqrt{59}}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{2\sqrt{59}}{3} from -6.

$$x_{3}=\frac{-\frac{2\sqrt{59}}{3}-6}{\frac{2}{3}}$$

Divide -6-\frac{2\sqrt{59}}{3} by \frac{2}{3} by multiplying -6-\frac{2\sqrt{59}}{3} by the reciprocal of \frac{2}{3}.

$$x_{3}=-\sqrt{59}-9$$

The equation is now solved.

$$x_{3}=\sqrt{59}-9 x_{3}=-\sqrt{59}-9$$

Steps for Completing the Square

Multiply 4 and \frac{1}{3} to get \frac{4}{3}.

$$\frac{4}{3}+6x_{3}+\frac{1}{3}x_{3}^{2}=-6$$

Subtract \frac{4}{3} from both sides.

$$6x_{3}+\frac{1}{3}x_{3}^{2}=-6-\frac{4}{3}$$

Subtract \frac{4}{3} from -6 to get -\frac{22}{3}.

$$6x_{3}+\frac{1}{3}x_{3}^{2}=-\frac{22}{3}$$

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.

$$\frac{1}{3}x_{3}^{2}+6x_{3}=-\frac{22}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x_{3}^{2}+6x_{3}}{\frac{1}{3}}=-\frac{\frac{22}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x_{3}^{2}+\frac{6}{\frac{1}{3}}x_{3}=-\frac{\frac{22}{3}}{\frac{1}{3}}$$

Divide 6 by \frac{1}{3} by multiplying 6 by the reciprocal of \frac{1}{3}.

$$x_{3}^{2}+18x_{3}=-\frac{\frac{22}{3}}{\frac{1}{3}}$$

Divide -\frac{22}{3} by \frac{1}{3} by multiplying -\frac{22}{3} by the reciprocal of \frac{1}{3}.

$$x_{3}^{2}+18x_{3}=-22$$

Divide 18, the coefficient of the x term, by 2 to get 9. Then add the square of 9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x_{3}^{2}+18x_{3}+9^{2}=-22+9^{2}$$

Square 9.

$$x_{3}^{2}+18x_{3}+81=-22+81$$

Add -22 to 81.

$$x_{3}^{2}+18x_{3}+81=59$$

Factor x_{3}^{2}+18x_{3}+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x_{3}+9\right)^{2}=59$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x_{3}+9\right)^{2}}=\sqrt{59}$$

Simplify.

$$x_{3}+9=\sqrt{59} x_{3}+9=-\sqrt{59}$$

Subtract 9 from both sides of the equation.

$$x_{3}=\sqrt{59}-9 x_{3}=-\sqrt{59}-9$$