Solve 1/x+2*(x+2)=3*(x+2)

$$\frac { 1 } { x + 2 } \cdot ( x + 2 ) = 3 \cdot ( x + 2 )$$

Solve for x

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$1\left(x+2\right)=3\left(x+2\right)\left(x+2\right)$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

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

Use the distributive property to multiply 1 by x+2.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x+2=3\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$x+2=3x^{2}+12x+12$$

Subtract 3x^{2} from both sides.

$$x+2-3x^{2}=12x+12$$

Subtract 12x from both sides.

$$x+2-3x^{2}-12x=12$$

Combine x and -12x to get -11x.

$$-11x+2-3x^{2}=12$$

Subtract 12 from both sides.

$$-11x+2-3x^{2}-12=0$$

Subtract 12 from 2 to get -10.

$$-11x-10-3x^{2}=0$$

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.

$$-3x^{2}-11x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -11 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-3\right)\left(-10\right)}}{2\left(-3\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-3\right)\left(-10\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-11\right)±\sqrt{121+12\left(-10\right)}}{2\left(-3\right)}$$

Multiply 12 times -10.

$$x=\frac{-\left(-11\right)±\sqrt{121-120}}{2\left(-3\right)}$$

Add 121 to -120.

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

Take the square root of 1.

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

The opposite of -11 is 11.

$$x=\frac{11±1}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{11±1}{-6}$$

Now solve the equation x=\frac{11±1}{-6} when ± is plus. Add 11 to 1.

$$x=\frac{12}{-6}$$

Divide 12 by -6.

$$x=-2$$

Now solve the equation x=\frac{11±1}{-6} when ± is minus. Subtract 1 from 11.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{3}$$

The equation is now solved.

$$x=-2 x=-\frac{5}{3}$$

Variable x cannot be equal to -2.

$$x=-\frac{5}{3}$$

Steps for Completing the Square

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$1\left(x+2\right)=3\left(x+2\right)\left(x+2\right)$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

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

Use the distributive property to multiply 1 by x+2.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x+2=3\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$x+2=3x^{2}+12x+12$$

Subtract 3x^{2} from both sides.

$$x+2-3x^{2}=12x+12$$

Subtract 12x from both sides.

$$x+2-3x^{2}-12x=12$$

Combine x and -12x to get -11x.

$$-11x+2-3x^{2}=12$$

Subtract 2 from both sides.

$$-11x-3x^{2}=12-2$$

Subtract 2 from 12 to get 10.

$$-11x-3x^{2}=10$$

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.

$$-3x^{2}-11x=10$$

Divide both sides by -3.

$$\frac{-3x^{2}-11x}{-3}=\frac{10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{11}{-3}\right)x=\frac{10}{-3}$$

Divide -11 by -3.

$$x^{2}+\frac{11}{3}x=\frac{10}{-3}$$

Divide 10 by -3.

$$x^{2}+\frac{11}{3}x=-\frac{10}{3}$$

Divide \frac{11}{3}, the coefficient of the x term, by 2 to get \frac{11}{6}. Then add the square of \frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=-\frac{10}{3}+\left(\frac{11}{6}\right)^{2}$$

Square \frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=-\frac{10}{3}+\frac{121}{36}$$

Add -\frac{10}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{11}{3}x+\frac{121}{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}.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{11}{6}=\frac{1}{6} x+\frac{11}{6}=-\frac{1}{6}$$

Subtract \frac{11}{6} from both sides of the equation.

$$x=-\frac{5}{3} x=-2$$

Variable x cannot be equal to -2.

$$x=-\frac{5}{3}$$