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

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

Solve for x

$$x=-3$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

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

Add 2x to both sides.

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

Combine \frac{1}{3}x and 2x to get \frac{7}{3}x.

$$\frac{7}{3}x+1+x^{2}=3$$

Subtract 3 from both sides.

$$\frac{7}{3}x+1+x^{2}-3=0$$

Subtract 3 from 1 to get -2.

$$\frac{7}{3}x-2+x^{2}=0$$

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.

$$x^{2}+\frac{7}{3}x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{7}{3} for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{7}{3}±\sqrt{\left(\frac{7}{3}\right)^{2}-4\left(-2\right)}}{2}$$

Square \frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{49}{9}-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{49}{9}+8}}{2}$$

Add \frac{49}{9} to 8.

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{121}{9}}}{2}$$

Take the square root of \frac{121}{9}.

$$x=\frac{-\frac{7}{3}±\frac{11}{3}}{2}$$

Now solve the equation x=\frac{-\frac{7}{3}±\frac{11}{3}}{2} when ± is plus. Add -\frac{7}{3} to \frac{11}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide \frac{4}{3} by 2.

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

Now solve the equation x=\frac{-\frac{7}{3}±\frac{11}{3}}{2} when ± is minus. Subtract \frac{11}{3} from -\frac{7}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

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

Add 2x to both sides.

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

Combine \frac{1}{3}x and 2x to get \frac{7}{3}x.

$$\frac{7}{3}x+1+x^{2}=3$$

Subtract 1 from both sides.

$$\frac{7}{3}x+x^{2}=3-1$$

Subtract 1 from 3 to get 2.

$$\frac{7}{3}x+x^{2}=2$$

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.

$$x^{2}+\frac{7}{3}x=2$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=2+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=2+\frac{49}{36}$$

Add 2 to \frac{49}{36}.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{121}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{11}{6} x+\frac{7}{6}=-\frac{11}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

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