Solve 1+3/x-2=12/x^2-4

$$1 + \frac { 3 } { x - 2 } = \frac { 12 } { x ^ { 2 } - 4 }$$

Solve for x

$$x=-5$$

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.

$$\left(x-2\right)\left(x+2\right)+\left(x+2\right)\times 3=12$$

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.

$$x^{2}-4+\left(x+2\right)\times 3=12$$

Use the distributive property to multiply x+2 by 3.

$$x^{2}-4+3x+6=12$$

Add -4 and 6 to get 2.

$$x^{2}+2+3x=12$$

Subtract 12 from both sides.

$$x^{2}+2+3x-12=0$$

Subtract 12 from 2 to get -10.

$$x^{2}-10+3x=0$$

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

$$x^{2}+3x-10=0$$

To solve the equation, factor x^{2}+3x-10 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.

$$a+b=3 ab=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

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

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

To find equation solutions, solve x-2=0 and x+5=0.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$

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.

$$\left(x-2\right)\left(x+2\right)+\left(x+2\right)\times 3=12$$

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.

$$x^{2}-4+\left(x+2\right)\times 3=12$$

Use the distributive property to multiply x+2 by 3.

$$x^{2}-4+3x+6=12$$

Add -4 and 6 to get 2.

$$x^{2}+2+3x=12$$

Subtract 12 from both sides.

$$x^{2}+2+3x-12=0$$

Subtract 12 from 2 to get -10.

$$x^{2}-10+3x=0$$

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

$$x^{2}+3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite x^{2}+3x-10 as \left(x^{2}-2x\right)+\left(5x-10\right).

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

Factor out x in the first and 5 in the second group.

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

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

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

To find equation solutions, solve x-2=0 and x+5=0.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$

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.

$$\left(x-2\right)\left(x+2\right)+\left(x+2\right)\times 3=12$$

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.

$$x^{2}-4+\left(x+2\right)\times 3=12$$

Use the distributive property to multiply x+2 by 3.

$$x^{2}-4+3x+6=12$$

Add -4 and 6 to get 2.

$$x^{2}+2+3x=12$$

Subtract 12 from both sides.

$$x^{2}+2+3x-12=0$$

Subtract 12 from 2 to get -10.

$$x^{2}-10+3x=0$$

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.

$$x^{2}+3x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$x=\frac{-3±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-3±7}{2}$$

Now solve the equation x=\frac{-3±7}{2} when ± is plus. Add -3 to 7.

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

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-3±7}{2} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$

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.

$$\left(x-2\right)\left(x+2\right)+\left(x+2\right)\times 3=12$$

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.

$$x^{2}-4+\left(x+2\right)\times 3=12$$

Use the distributive property to multiply x+2 by 3.

$$x^{2}-4+3x+6=12$$

Add -4 and 6 to get 2.

$$x^{2}+2+3x=12$$

Subtract 2 from both sides.

$$x^{2}+3x=12-2$$

Subtract 2 from 12 to get 10.

$$x^{2}+3x=10$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{49}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{7}{2} x+\frac{3}{2}=-\frac{7}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$