Solve x+1/3+5/x+1+2=0

$$\frac{ x+1 }{ 3 } + \frac{ 5 }{ x+1 } +2=0$$

Solve for x (complex solution)

$$x=-\sqrt{6}i-4\approx -4-2.449489743i$$

$$x=-4+\sqrt{6}i\approx -4+2.449489743i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 3\left(x+1\right), the least common multiple of 3,x+1.

$$\left(x+1\right)\left(x+1\right)+3\times 5+3\left(x+1\right)\times 2=0$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\left(x+1\right)^{2}+3\times 5+3\left(x+1\right)\times 2=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1+3\times 5+3\left(x+1\right)\times 2=0$$

Multiply 3 and 5 to get 15.

$$x^{2}+2x+1+15+3\left(x+1\right)\times 2=0$$

Add 1 and 15 to get 16.

$$x^{2}+2x+16+3\left(x+1\right)\times 2=0$$

Multiply 3 and 2 to get 6.

$$x^{2}+2x+16+6\left(x+1\right)=0$$

Use the distributive property to multiply 6 by x+1.

$$x^{2}+2x+16+6x+6=0$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x+16+6=0$$

Add 16 and 6 to get 22.

$$x^{2}+8x+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 22}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 22}}{2}$$

Multiply -4 times 22.

$$x=\frac{-8±\sqrt{64-88}}{2}$$

Add 64 to -88.

$$x=\frac{-8±\sqrt{-24}}{2}$$

Take the square root of -24.

$$x=\frac{-8±2\sqrt{6}i}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{6}i}{2} when ± is plus. Add -8 to 2i\sqrt{6}.

$$x=\frac{-8+2\sqrt{6}i}{2}$$

Divide -8+2i\sqrt{6} by 2.

$$x=-4+\sqrt{6}i$$

Now solve the equation x=\frac{-8±2\sqrt{6}i}{2} when ± is minus. Subtract 2i\sqrt{6} from -8.

$$x=\frac{-2\sqrt{6}i-8}{2}$$

Divide -8-2i\sqrt{6} by 2.

$$x=-\sqrt{6}i-4$$

The equation is now solved.

$$x=-4+\sqrt{6}i x=-\sqrt{6}i-4$$

Steps for Completing the Square

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 3\left(x+1\right), the least common multiple of 3,x+1.

$$\left(x+1\right)\left(x+1\right)+3\times 5+3\left(x+1\right)\times 2=0$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\left(x+1\right)^{2}+3\times 5+3\left(x+1\right)\times 2=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1+3\times 5+3\left(x+1\right)\times 2=0$$

Multiply 3 and 5 to get 15.

$$x^{2}+2x+1+15+3\left(x+1\right)\times 2=0$$

Add 1 and 15 to get 16.

$$x^{2}+2x+16+3\left(x+1\right)\times 2=0$$

Multiply 3 and 2 to get 6.

$$x^{2}+2x+16+6\left(x+1\right)=0$$

Use the distributive property to multiply 6 by x+1.

$$x^{2}+2x+16+6x+6=0$$

Combine 2x and 6x to get 8x.

$$x^{2}+8x+16+6=0$$

Add 16 and 6 to get 22.

$$x^{2}+8x+22=0$$

Subtract 22 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+8x=-22$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-22+4^{2}$$

Square 4.

$$x^{2}+8x+16=-22+16$$

Add -22 to 16.

$$x^{2}+8x+16=-6$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{-6}$$

Simplify.

$$x+4=\sqrt{6}i x+4=-\sqrt{6}i$$

Subtract 4 from both sides of the equation.

$$x=-4+\sqrt{6}i x=-\sqrt{6}i-4$$