Solve 1-5/x-2=x+2/x^2-4

$$1 - \frac { 5 } { x - 2 } = \frac { x + 2 } { x ^ { 2 } - 4 }$$

Solve for x

$$x=8$$

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

$$\left(x-2\right)\left(x+2\right)-\left(x+2\right)\times 5=x+2$$

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.

$$x^{2}-4-\left(x+2\right)\times 5=x+2$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-4-\left(5x+10\right)=x+2$$

To find the opposite of 5x+10, find the opposite of each term.

$$x^{2}-4-5x-10=x+2$$

Subtract 10 from -4 to get -14.

$$x^{2}-14-5x=x+2$$

Subtract x from both sides.

$$x^{2}-14-5x-x=2$$

Combine -5x and -x to get -6x.

$$x^{2}-14-6x=2$$

Subtract 2 from both sides.

$$x^{2}-14-6x-2=0$$

Subtract 2 from -14 to get -16.

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

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

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

To solve the equation, factor x^{2}-6x-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.

$$a+b=-6 ab=-16$$

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 -16.

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

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$a=-8 b=2$$

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

$$\left(x-8\right)\left(x+2\right)$$

To find equation solutions, solve x-8=0 and x+2=0.

$$x=8 x=-2$$

Variable x cannot be equal to -2.

$$x=8$$

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.

$$\left(x-2\right)\left(x+2\right)-\left(x+2\right)\times 5=x+2$$

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.

$$x^{2}-4-\left(x+2\right)\times 5=x+2$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-4-\left(5x+10\right)=x+2$$

To find the opposite of 5x+10, find the opposite of each term.

$$x^{2}-4-5x-10=x+2$$

Subtract 10 from -4 to get -14.

$$x^{2}-14-5x=x+2$$

Subtract x from both sides.

$$x^{2}-14-5x-x=2$$

Combine -5x and -x to get -6x.

$$x^{2}-14-6x=2$$

Subtract 2 from both sides.

$$x^{2}-14-6x-2=0$$

Subtract 2 from -14 to get -16.

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

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

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

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.

$$a+b=-6 ab=1\left(-16\right)=-16$$

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 -16.

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

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$a=-8 b=2$$

Rewrite x^{2}-6x-16 as \left(x^{2}-8x\right)+\left(2x-16\right).

$$\left(x^{2}-8x\right)+\left(2x-16\right)$$

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

$$x\left(x-8\right)+2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+2\right)$$

To find equation solutions, solve x-8=0 and x+2=0.

$$x=8 x=-2$$

Variable x cannot be equal to -2.

$$x=8$$

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.

$$\left(x-2\right)\left(x+2\right)-\left(x+2\right)\times 5=x+2$$

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.

$$x^{2}-4-\left(x+2\right)\times 5=x+2$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-4-\left(5x+10\right)=x+2$$

To find the opposite of 5x+10, find the opposite of each term.

$$x^{2}-4-5x-10=x+2$$

Subtract 10 from -4 to get -14.

$$x^{2}-14-5x=x+2$$

Subtract x from both sides.

$$x^{2}-14-5x-x=2$$

Combine -5x and -x to get -6x.

$$x^{2}-14-6x=2$$

Subtract 2 from both sides.

$$x^{2}-14-6x-2=0$$

Subtract 2 from -14 to get -16.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-16\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-\left(-6\right)±\sqrt{36+64}}{2}$$

Add 36 to 64.

$$x=\frac{-\left(-6\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-\left(-6\right)±10}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±10}{2}$$

Now solve the equation x=\frac{6±10}{2} when ± is plus. Add 6 to 10.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{6±10}{2} when ± is minus. Subtract 10 from 6.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=8 x=-2$$

Variable x cannot be equal to -2.

$$x=8$$

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.

$$\left(x-2\right)\left(x+2\right)-\left(x+2\right)\times 5=x+2$$

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.

$$x^{2}-4-\left(x+2\right)\times 5=x+2$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-4-\left(5x+10\right)=x+2$$

To find the opposite of 5x+10, find the opposite of each term.

$$x^{2}-4-5x-10=x+2$$

Subtract 10 from -4 to get -14.

$$x^{2}-14-5x=x+2$$

Subtract x from both sides.

$$x^{2}-14-5x-x=2$$

Combine -5x and -x to get -6x.

$$x^{2}-14-6x=2$$

Add 14 to both sides.

$$x^{2}-6x=2+14$$

Add 2 and 14 to get 16.

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=16+9$$

Add 16 to 9.

$$x^{2}-6x+9=25$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-3=5 x-3=-5$$

Add 3 to both sides of the equation.

$$x=8 x=-2$$

Variable x cannot be equal to -2.

$$x=8$$