Solve 1/x+1/x+2+1/x+4=0

$$\frac{ 1 }{ x } + \frac{ 1 }{ x+2 } + \frac{ 1 }{ x+4 } =0$$

Solve for x

$$x=\frac{2\sqrt{3}}{3}-2\approx -0.845299462$$

$$x=-\frac{2\sqrt{3}}{3}-2\approx -3.154700538$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,-2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right)\left(x+4\right), the least common multiple of x,x+2,x+4.

$$\left(x+2\right)\left(x+4\right)+x\left(x+4\right)+x\left(x+2\right)=0$$

Use the distributive property to multiply x+2 by x+4 and combine like terms.

$$x^{2}+6x+8+x\left(x+4\right)+x\left(x+2\right)=0$$

Use the distributive property to multiply x by x+4.

$$x^{2}+6x+8+x^{2}+4x+x\left(x+2\right)=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+6x+8+4x+x\left(x+2\right)=0$$

Combine 6x and 4x to get 10x.

$$2x^{2}+10x+8+x\left(x+2\right)=0$$

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

$$2x^{2}+10x+8+x^{2}+2x=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+10x+8+2x=0$$

Combine 10x and 2x to get 12x.

$$3x^{2}+12x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 12 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\times 8}}{2\times 3}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-12±\sqrt{144-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$x=\frac{-12±\sqrt{144-96}}{2\times 3}$$

Add 144 to -96.

$$x=\frac{-12±\sqrt{48}}{2\times 3}$$

Take the square root of 48.

$$x=\frac{-12±4\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-12±4\sqrt{3}}{6}$$

Now solve the equation x=\frac{-12±4\sqrt{3}}{6} when ± is plus. Add -12 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-12}{6}$$

Divide -12+4\sqrt{3} by 6.

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

Now solve the equation x=\frac{-12±4\sqrt{3}}{6} when ± is minus. Subtract 4\sqrt{3} from -12.

$$x=\frac{-4\sqrt{3}-12}{6}$$

Divide -12-4\sqrt{3} by 6.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,-2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right)\left(x+4\right), the least common multiple of x,x+2,x+4.

$$\left(x+2\right)\left(x+4\right)+x\left(x+4\right)+x\left(x+2\right)=0$$

Use the distributive property to multiply x+2 by x+4 and combine like terms.

$$x^{2}+6x+8+x\left(x+4\right)+x\left(x+2\right)=0$$

Use the distributive property to multiply x by x+4.

$$x^{2}+6x+8+x^{2}+4x+x\left(x+2\right)=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+6x+8+4x+x\left(x+2\right)=0$$

Combine 6x and 4x to get 10x.

$$2x^{2}+10x+8+x\left(x+2\right)=0$$

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

$$2x^{2}+10x+8+x^{2}+2x=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+10x+8+2x=0$$

Combine 10x and 2x to get 12x.

$$3x^{2}+12x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+12x=-8$$

Divide both sides by 3.

$$\frac{3x^{2}+12x}{3}=-\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{12}{3}x=-\frac{8}{3}$$

Divide 12 by 3.

$$x^{2}+4x=-\frac{8}{3}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=-\frac{8}{3}+2^{2}$$

Square 2.

$$x^{2}+4x+4=-\frac{8}{3}+4$$

Add -\frac{8}{3} to 4.

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

Factor x^{2}+4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 2 from both sides of the equation.

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