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

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

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.

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

Add -2 and 4 to get 2.

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

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.

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

Subtract x^{2} from both sides.

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

Add 4 to both sides.

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

Add 2 and 4 to get 6.

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

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.

$$a+b=1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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.

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

Add -2 and 4 to get 2.

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

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.

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

Subtract x^{2} from both sides.

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

Add 4 to both sides.

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

Add 2 and 4 to get 6.

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

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-1±\sqrt{1+24}}{2\left(-1\right)}$$

Add 1 to 24.

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

Take the square root of 25.

$$x=\frac{-1±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±5}{-2}$$

Now solve the equation x=\frac{-1±5}{-2} when ± is plus. Add -1 to 5.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-1±5}{-2} when ± is minus. Subtract 5 from -1.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-2 x=3$$

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.

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

Add -2 and 4 to get 2.

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

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.

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

Subtract x^{2} from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from -4 to get -6.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -6 by -1.

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

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.

$$x=3$$