Solve 6/n-4+6/n+4=2

$$\frac { 6 } { n - 4 } + \frac { 6 } { n + 4 } = 2$$

Solve for n

$$n=-2$$

$$n=8$$

Steps Using the Quadratic Formula

Variable n cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(n-4\right)\left(n+4\right), the least common multiple of n-4,n+4.

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

Use the distributive property to multiply n+4 by 6.

$$6n+24+\left(n-4\right)\times 6=2\left(n-4\right)\left(n+4\right)$$

Use the distributive property to multiply n-4 by 6.

$$6n+24+6n-24=2\left(n-4\right)\left(n+4\right)$$

Combine 6n and 6n to get 12n.

$$12n+24-24=2\left(n-4\right)\left(n+4\right)$$

Subtract 24 from 24 to get 0.

$$12n=2\left(n-4\right)\left(n+4\right)$$

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

$$12n=\left(2n-8\right)\left(n+4\right)$$

Use the distributive property to multiply 2n-8 by n+4 and combine like terms.

$$12n=2n^{2}-32$$

Subtract 2n^{2} from both sides.

$$12n-2n^{2}=-32$$

Add 32 to both sides.

$$12n-2n^{2}+32=0$$

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.

$$-2n^{2}+12n+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

Square 12.

$$n=\frac{-12±\sqrt{144-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$n=\frac{-12±\sqrt{144+8\times 32}}{2\left(-2\right)}$$

Multiply 8 times 32.

$$n=\frac{-12±\sqrt{144+256}}{2\left(-2\right)}$$

Add 144 to 256.

$$n=\frac{-12±\sqrt{400}}{2\left(-2\right)}$$

Take the square root of 400.

$$n=\frac{-12±20}{2\left(-2\right)}$$

Multiply 2 times -2.

$$n=\frac{-12±20}{-4}$$

Now solve the equation n=\frac{-12±20}{-4} when ± is plus. Add -12 to 20.

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

Divide 8 by -4.

$$n=-2$$

Now solve the equation n=\frac{-12±20}{-4} when ± is minus. Subtract 20 from -12.

$$n=-\frac{32}{-4}$$

Divide -32 by -4.

$$n=8$$

The equation is now solved.

$$n=-2 n=8$$

Steps for Completing the Square

Variable n cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(n-4\right)\left(n+4\right), the least common multiple of n-4,n+4.

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

Use the distributive property to multiply n+4 by 6.

$$6n+24+\left(n-4\right)\times 6=2\left(n-4\right)\left(n+4\right)$$

Use the distributive property to multiply n-4 by 6.

$$6n+24+6n-24=2\left(n-4\right)\left(n+4\right)$$

Combine 6n and 6n to get 12n.

$$12n+24-24=2\left(n-4\right)\left(n+4\right)$$

Subtract 24 from 24 to get 0.

$$12n=2\left(n-4\right)\left(n+4\right)$$

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

$$12n=\left(2n-8\right)\left(n+4\right)$$

Use the distributive property to multiply 2n-8 by n+4 and combine like terms.

$$12n=2n^{2}-32$$

Subtract 2n^{2} from both sides.

$$12n-2n^{2}=-32$$

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.

$$-2n^{2}+12n=-32$$

Divide both sides by -2.

$$\frac{-2n^{2}+12n}{-2}=-\frac{32}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$n^{2}+\frac{12}{-2}n=-\frac{32}{-2}$$

Divide 12 by -2.

$$n^{2}-6n=-\frac{32}{-2}$$

Divide -32 by -2.

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

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.

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

Square -3.

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

Add 16 to 9.

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

Factor n^{2}-6n+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$n-3=5 n-3=-5$$

Add 3 to both sides of the equation.

$$n=8 n=-2$$