Solve x+2/x=x^2+5x+4/x^2+2x+x/x+2

$$\frac { x + 2 } { x } = \frac { x ^ { 2 } + 5 x + 4 } { x ^ { 2 } + 2 x } + \frac { x } { x + 2 }$$

Solve for x

$$x=-1$$

Steps Using Factoring

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

$$\left(x+2\right)\left(x+2\right)=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=2x^{2}+5x+4$$

Subtract 2x^{2} from both sides.

$$x^{2}+4x+4-2x^{2}=5x+4$$

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

$$-x^{2}+4x+4=5x+4$$

Subtract 5x from both sides.

$$-x^{2}+4x+4-5x=4$$

Combine 4x and -5x to get -x.

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

Subtract 4 from both sides.

$$-x^{2}-x+4-4=0$$

Subtract 4 from 4 to get 0.

$$-x^{2}-x=0$$

Factor out x.

$$x\left(-x-1\right)=0$$

To find equation solutions, solve x=0 and -x-1=0.

$$x=0 x=-1$$

Variable x cannot be equal to 0.

$$x=-1$$

Steps Using the Quadratic Formula

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

$$\left(x+2\right)\left(x+2\right)=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=2x^{2}+5x+4$$

Subtract 2x^{2} from both sides.

$$x^{2}+4x+4-2x^{2}=5x+4$$

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

$$-x^{2}+4x+4=5x+4$$

Subtract 5x from both sides.

$$-x^{2}+4x+4-5x=4$$

Combine 4x and -5x to get -x.

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

Subtract 4 from both sides.

$$-x^{2}-x+4-4=0$$

Subtract 4 from 4 to get 0.

$$-x^{2}-x=0$$

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±1}{-2}$$

Now solve the equation x=\frac{1±1}{-2} when ± is plus. Add 1 to 1.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{1±1}{-2} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-1 x=0$$

Variable x cannot be equal to 0.

$$x=-1$$

Steps for Completing the Square

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

$$\left(x+2\right)\left(x+2\right)=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+xx$$

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

$$\left(x+2\right)^{2}=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=x^{2}+5x+4+x^{2}$$

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

$$x^{2}+4x+4=2x^{2}+5x+4$$

Subtract 2x^{2} from both sides.

$$x^{2}+4x+4-2x^{2}=5x+4$$

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

$$-x^{2}+4x+4=5x+4$$

Subtract 5x from both sides.

$$-x^{2}+4x+4-5x=4$$

Combine 4x and -5x to get -x.

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

Subtract 4 from both sides.

$$-x^{2}-x=4-4$$

Subtract 4 from 4 to get 0.

$$-x^{2}-x=0$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=\frac{0}{-1}$$

Divide -1 by -1.

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

Divide 0 by -1.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

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

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

Factor x^{2}+x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=0 x=-1$$

Variable x cannot be equal to 0.

$$x=-1$$