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

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

Solve for x

$$x=0$$

Steps Using Factoring

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

$$\left(x+2\right)^{2}=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x^{2}-4x+4\right)\times 0-\left(x^{2}-4\right)$$

Anything times zero gives zero.

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

To find the opposite of x^{2}-4, find the opposite of each term.

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

Add 0 and 4 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add x^{2} to both sides.

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

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

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

Factor out x.

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

To find equation solutions, solve x=0 and 2x+4=0.

$$x=0 x=-2$$

Variable x cannot be equal to -2.

$$x=0$$

Steps Using the Quadratic Formula

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

$$\left(x+2\right)^{2}=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x^{2}-4x+4\right)\times 0-\left(x^{2}-4\right)$$

Anything times zero gives zero.

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

To find the opposite of x^{2}-4, find the opposite of each term.

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

Add 0 and 4 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add x^{2} to both sides.

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

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

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

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

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±4}{4}$$

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

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

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

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Variable x cannot be equal to -2.

$$x=0$$

Steps for Completing the Square

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

$$\left(x+2\right)^{2}=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x-2\right)^{2}\times 0-\left(x^{2}-4\right)$$

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

$$x^{2}+4x+4=\left(x^{2}-4x+4\right)\times 0-\left(x^{2}-4\right)$$

Anything times zero gives zero.

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

To find the opposite of x^{2}-4, find the opposite of each term.

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

Add 0 and 4 to get 4.

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

Add x^{2} to both sides.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

Divide 0 by 2.

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

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

$$x^{2}+2x+1^{2}=1^{2}$$

Square 1.

$$x^{2}+2x+1=1$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$

Variable x cannot be equal to -2.

$$x=0$$