Solve 5/x^2-4+3/x+2=4/x-2-3/1

$$\frac { 5 } { x ^ { 2 } - 4 } + \frac { 3 } { x + 2 } = \frac { 4 } { x - 2 } - \frac { 3 } { 1 }$$

Solve for x

$$x = \frac{\sqrt{253} + 1}{6} \approx 2.817662287$$

$$x=\frac{1-\sqrt{253}}{6}\approx -2.484328953$$

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}-4,x+2,x-2.

$$5+\left(x-2\right)\times 3=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

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

$$5+3x-6=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

Subtract 6 from 5 to get -1.

$$-1+3x=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

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

$$-1+3x=4x+8-\left(x^{2}-4\right)\times 3$$

Use the distributive property to multiply x^{2}-4 by 3.

$$-1+3x=4x+8-\left(3x^{2}-12\right)$$

To find the opposite of 3x^{2}-12, find the opposite of each term.

$$-1+3x=4x+8-3x^{2}+12$$

Add 8 and 12 to get 20.

$$-1+3x=4x+20-3x^{2}$$

Subtract 4x from both sides.

$$-1+3x-4x=20-3x^{2}$$

Combine 3x and -4x to get -x.

$$-1-x=20-3x^{2}$$

Subtract 20 from both sides.

$$-1-x-20=-3x^{2}$$

Subtract 20 from -1 to get -21.

$$-21-x=-3x^{2}$$

Add 3x^{2} to both sides.

$$-21-x+3x^{2}=0$$

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.

$$3x^{2}-x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-21\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-21\right)}}{2\times 3}$$

Multiply -12 times -21.

$$x=\frac{-\left(-1\right)±\sqrt{1+252}}{2\times 3}$$

Add 1 to 252.

$$x=\frac{-\left(-1\right)±\sqrt{253}}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{253}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±\sqrt{253}}{6}$$

Now solve the equation x=\frac{1±\sqrt{253}}{6} when ± is plus. Add 1 to \sqrt{253}.

$$x=\frac{\sqrt{253}+1}{6}$$

Now solve the equation x=\frac{1±\sqrt{253}}{6} when ± is minus. Subtract \sqrt{253} from 1.

$$x=\frac{1-\sqrt{253}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{253}+1}{6} x=\frac{1-\sqrt{253}}{6}$$

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}-4,x+2,x-2.

$$5+\left(x-2\right)\times 3=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

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

$$5+3x-6=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

Subtract 6 from 5 to get -1.

$$-1+3x=\left(x+2\right)\times 4-\left(x^{2}-4\right)\times 3$$

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

$$-1+3x=4x+8-\left(x^{2}-4\right)\times 3$$

Use the distributive property to multiply x^{2}-4 by 3.

$$-1+3x=4x+8-\left(3x^{2}-12\right)$$

To find the opposite of 3x^{2}-12, find the opposite of each term.

$$-1+3x=4x+8-3x^{2}+12$$

Add 8 and 12 to get 20.

$$-1+3x=4x+20-3x^{2}$$

Subtract 4x from both sides.

$$-1+3x-4x=20-3x^{2}$$

Combine 3x and -4x to get -x.

$$-1-x=20-3x^{2}$$

Add 3x^{2} to both sides.

$$-1-x+3x^{2}=20$$

Add 1 to both sides.

$$-x+3x^{2}=20+1$$

Add 20 and 1 to get 21.

$$-x+3x^{2}=21$$

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.

$$3x^{2}-x=21$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{21}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{21}{3}$$

Divide 21 by 3.

$$x^{2}-\frac{1}{3}x=7$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=7+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=7+\frac{1}{36}$$

Add 7 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{253}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{253}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{253}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{253}}{6} x-\frac{1}{6}=-\frac{\sqrt{253}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{253}+1}{6} x=\frac{1-\sqrt{253}}{6}$$