Solve x=1/6(x+1)(x+2)

$$x = \frac { 1 } { 6 } ( x + 1 ) ( x + 2 )$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{6} by x+1.

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

Use the distributive property to multiply \frac{1}{6}x+\frac{1}{6} by x+2 and combine like terms.

$$x=\frac{1}{6}x^{2}+\frac{1}{2}x+\frac{1}{3}$$

Subtract \frac{1}{6}x^{2} from both sides.

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

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

$$x-\frac{1}{6}x^{2}-\frac{1}{2}x=\frac{1}{3}$$

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

$$\frac{1}{2}x-\frac{1}{6}x^{2}=\frac{1}{3}$$

Subtract \frac{1}{3} from both sides.

$$\frac{1}{2}x-\frac{1}{6}x^{2}-\frac{1}{3}=0$$

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.

$$-\frac{1}{6}x^{2}+\frac{1}{2}x-\frac{1}{3}=0$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-\frac{1}{6}\right)\left(-\frac{1}{3}\right)}}{2\left(-\frac{1}{6}\right)}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-\frac{1}{6}\right)\left(-\frac{1}{3}\right)}}{2\left(-\frac{1}{6}\right)}$$

Multiply -4 times -\frac{1}{6}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+\frac{2}{3}\left(-\frac{1}{3}\right)}}{2\left(-\frac{1}{6}\right)}$$

Multiply \frac{2}{3} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-\frac{2}{9}}}{2\left(-\frac{1}{6}\right)}$$

Add \frac{1}{4} to -\frac{2}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Take the square root of \frac{1}{36}.

$$x=\frac{-\frac{1}{2}±\frac{1}{6}}{2\left(-\frac{1}{6}\right)}$$

Multiply 2 times -\frac{1}{6}.

$$x=\frac{-\frac{1}{2}±\frac{1}{6}}{-\frac{1}{3}}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{1}{6}}{-\frac{1}{3}} when ± is plus. Add -\frac{1}{2} to \frac{1}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=1$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{1}{6}}{-\frac{1}{3}} when ± is minus. Subtract \frac{1}{6} from -\frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{6} by x+1.

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

Use the distributive property to multiply \frac{1}{6}x+\frac{1}{6} by x+2 and combine like terms.

$$x=\frac{1}{6}x^{2}+\frac{1}{2}x+\frac{1}{3}$$

Subtract \frac{1}{6}x^{2} from both sides.

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

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

$$x-\frac{1}{6}x^{2}-\frac{1}{2}x=\frac{1}{3}$$

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

$$\frac{1}{2}x-\frac{1}{6}x^{2}=\frac{1}{3}$$

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.

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

Multiply both sides by -6.

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

Dividing by -\frac{1}{6} undoes the multiplication by -\frac{1}{6}.

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

Divide \frac{1}{2} by -\frac{1}{6} by multiplying \frac{1}{2} by the reciprocal of -\frac{1}{6}.

$$x^{2}-3x=\frac{\frac{1}{3}}{-\frac{1}{6}}$$

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

$$x^{2}-3x=-2$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$