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

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

Solve for x

$$x=-2$$

Steps Using Factoring

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

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

Use the distributive property to multiply x^{2}+2 by 2.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 4x to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor x^{2}+4x+4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

Rewrite as a binomial square.

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

To find equation solution, solve x+2=0.

$$x=-2$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply x^{2}+2 by 2.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 4x to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite x^{2}+4x+4 as \left(x^{2}+2x\right)+\left(2x+4\right).

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

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

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

Factor out common term x+2 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve x+2=0.

$$x=-2$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x^{2}+2 by 2.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 4x to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16}}{2}$$

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x^{2}+2 by 2.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 4x to both sides.

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

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

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

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.

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

Square 2.

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

Add -4 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=0 x+2=0$$

Subtract 2 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$x=-2$$