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

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

Solve for x

$$x=3$$

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

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

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Use the distributive property to multiply x^{2}-4 by 2.

$$2x^{2}-8+\left(x-2\right)\times 4=6x-4$$

Use the distributive property to multiply x-2 by 4.

$$2x^{2}-8+4x-8=6x-4$$

Subtract 8 from -8 to get -16.

$$2x^{2}-16+4x=6x-4$$

Subtract 6x from both sides.

$$2x^{2}-16+4x-6x=-4$$

Combine 4x and -6x to get -2x.

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

Add 4 to both sides.

$$2x^{2}-16-2x+4=0$$

Add -16 and 4 to get -12.

$$2x^{2}-12-2x=0$$

Divide both sides by 2.

$$x^{2}-6-x=0$$

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

$$x^{2}-x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite x^{2}-x-6 as \left(x^{2}-3x\right)+\left(2x-6\right).

$$\left(x^{2}-3x\right)+\left(2x-6\right)$$

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

$$x\left(x-3\right)+2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Variable x cannot be equal to -2.

$$x=3$$

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

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

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Use the distributive property to multiply x^{2}-4 by 2.

$$2x^{2}-8+\left(x-2\right)\times 4=6x-4$$

Use the distributive property to multiply x-2 by 4.

$$2x^{2}-8+4x-8=6x-4$$

Subtract 8 from -8 to get -16.

$$2x^{2}-16+4x=6x-4$$

Subtract 6x from both sides.

$$2x^{2}-16+4x-6x=-4$$

Combine 4x and -6x to get -2x.

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

Add 4 to both sides.

$$2x^{2}-16-2x+4=0$$

Add -16 and 4 to get -12.

$$2x^{2}-12-2x=0$$

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.

$$2x^{2}-2x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 2}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±10}{4}$$

Now solve the equation x=\frac{2±10}{4} when ± is plus. Add 2 to 10.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{2±10}{4} when ± is minus. Subtract 10 from 2.

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=3 x=-2$$

Variable x cannot be equal to -2.

$$x=3$$

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

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

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Use the distributive property to multiply x^{2}-4 by 2.

$$2x^{2}-8+\left(x-2\right)\times 4=6x-4$$

Use the distributive property to multiply x-2 by 4.

$$2x^{2}-8+4x-8=6x-4$$

Subtract 8 from -8 to get -16.

$$2x^{2}-16+4x=6x-4$$

Subtract 6x from both sides.

$$2x^{2}-16+4x-6x=-4$$

Combine 4x and -6x to get -2x.

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

Add 16 to both sides.

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

Add -4 and 16 to get 12.

$$2x^{2}-2x=12$$

Divide both sides by 2.

$$\frac{2x^{2}-2x}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

$$x^{2}-x=\frac{12}{2}$$

Divide 12 by 2.

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

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

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

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

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$

Variable x cannot be equal to -2.

$$x=3$$