Solve {l}{1/3x-1=x^2-2x-3}hline

$$\left. \begin{array} { l } { \frac { 1 } { 3 } x - 1 = x ^ { 2 } - 2 x - 3 } \\ \hline \end{array} \right.$$

Solve for x

$$x=3$$

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

Steps Using the Quadratic Formula

Subtract x^{2} from 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.

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

Add 3 to both sides.

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

Add -1 and 3 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}.

$$x=\frac{-\frac{7}{3}±\sqrt{\left(\frac{7}{3}\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

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

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{49}{9}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{49}{9}+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{49}{9}+8}}{2\left(-1\right)}$$

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

$$x=\frac{-\frac{7}{3}±\sqrt{\frac{121}{9}}}{2\left(-1\right)}$$

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

$$x=\frac{-\frac{7}{3}±\frac{11}{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

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

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

Subtract x^{2} from 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.

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

Add 1 to both sides.

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

Add -3 and 1 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 both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide \frac{7}{3} by -1.

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

Divide -2 by -1.

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

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.

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

Add \frac{7}{6} to both sides of the equation.

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