Solve 6/x^2-4-x+1/2-x=x/2x+4

$$\frac { 6 } { x ^ { 2 } - 4 } - \frac { x + 1 } { 2 - x } = \frac { x } { 2 x + 4 }$$

Solve for x

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Multiply 2 and 6 to get 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Use the distributive property to multiply -4-2x by x+1 and combine like terms.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

To find the opposite of -6x-4-2x^{2}, find the opposite of each term.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Add 12 and 4 to get 16.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 6x and 2x to get 8x.

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

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

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

To solve the equation, factor x^{2}+8x+16 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.

$$a+b=8 ab=16$$

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 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

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

$$\left(x+4\right)\left(x+4\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x+4=0.

$$x=-4$$

Steps Using Factoring By Grouping

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 2\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Multiply 2 and 6 to get 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Use the distributive property to multiply -4-2x by x+1 and combine like terms.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

To find the opposite of -6x-4-2x^{2}, find the opposite of each term.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Add 12 and 4 to get 16.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 6x and 2x to get 8x.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\times 16=16$$

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 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite x^{2}+8x+16 as \left(x^{2}+4x\right)+\left(4x+16\right).

$$\left(x^{2}+4x\right)+\left(4x+16\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+4\right)+4\left(x+4\right)$$

Factor out common term x+4 by using distributive property.

$$\left(x+4\right)\left(x+4\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x+4=0.

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Multiply 2 and 6 to get 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Use the distributive property to multiply -4-2x by x+1 and combine like terms.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

To find the opposite of -6x-4-2x^{2}, find the opposite of each term.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Add 12 and 4 to get 16.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 6x and 2x to get 8x.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16}}{2}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64}}{2}$$

Add 64 to -64.

$$x=\frac{-8±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Multiply 2 and 6 to get 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Use the distributive property to multiply -4-2x by x+1 and combine like terms.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

To find the opposite of -6x-4-2x^{2}, find the opposite of each term.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Add 12 and 4 to get 16.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 6x and 2x to get 8x.

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

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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.

$$x^{2}+8x+4^{2}=-16+4^{2}$$

Square 4.

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

Add -16 to 16.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=0 x+4=0$$

Subtract 4 from both sides of the equation.

$$x=-4 x=-4$$

The equation is now solved. Solutions are the same.

$$x=-4$$