Solve x^2+4x+4/x^2+2x=-1

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

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).

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

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

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

Add x^{2} to both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+4=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite x^{2}+3x+2 as \left(x^{2}+x\right)+\left(2x+2\right).

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

Factor out x in the first and 2 in the second group.

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

Factor out common term x+1 by using distributive property.

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

To find equation solutions, solve x+1=0 and x+2=0.

$$x=-1 x=-2$$

Variable x cannot be equal to -2.

$$x=-1$$

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).

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

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

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

Add x^{2} to both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+4=0$$

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

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\times 4}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-6±\sqrt{36-32}}{2\times 2}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$x=\frac{-6±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±2}{4}$$

Now solve the equation x=\frac{-6±2}{4} when ± is plus. Add -6 to 2.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-6±2}{4} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=-1 x=-2$$

Variable x cannot be equal to -2.

$$x=-1$$

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).

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

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

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

Add x^{2} to both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+6x=-4$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=-\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=-\frac{4}{2}$$

Divide 6 by 2.

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

Divide -4 by 2.

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

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.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=-1 x=-2$$

Variable x cannot be equal to -2.

$$x=-1$$