# Solve the following equation: 1/(x-1) + 1/(x+1) = 18

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We have to solve 1/(x-1) + 1/(x+1) = 18

1/(x-1) + 1/(x+1) = 18

Multiply all the terms by (x-1) (x+1) to eliminate the denominators.

=> (x+1) + (x-1) = 18(x-1) (x+1)

=> 2x = 18 (x^2 – 1)

Divide both the sides by 2

=> x = 9(x^2 – 1)

=> x = 9x^2 – 9

=> 9x^2 – x – 9 =0

Now we have a quadratic equation that can be solved by using the following relation x1 = [-b –sqrt (b^2 – 4ax)]/2a and x2 = [-b –sqrt (b^2 – 4ax)]/2a

now sqrt (b^2 – 4ac) = sqrt (1 + 4*4*9) = sqrt (325)

x1 = 1/18 – (sqrt 325)/18

x2 = 1/18 + (sqrt 325)/18

**The roots of the equation are 1/18 – (sqrt 325)/18 and 1/18 + (sqrt 325)/18.**

First, we'll move all terms to one side:

1/(x-1) + 1/(x+1) - 18 = 0

Now, we'll calculate the least common denominator for adding the ratios:

LCD = (x-1)(x+1)

We notice that the result of the product is the difference of squares:

(x-1)(x+1) = x^2 - 1

We'll re-write the equation:

x + 1 + x - 1 - 18(x^2 - 1) = 0

We'll remove the brackets and we'll combine and eliminate like terms:

2x - 18x^2 + 18 = 0

We'll divide by -2 and we'll re-arrange the terms:

9x^2 - x - 9 = 0

We'll apply the quadratic formula:

x1 = [1 + sqrt(1 + 324)]/18

**x1 = (1+5sqrt13)/18**

**x2 = (1-5sqrt13)/18**

**Since the roots are different from the values 1 and -1, we'll accept them.**

The roots of a quadratic equation are given by [-b + sqrt(b^2 - 4ac)]/ 2a and [-b - sqrt(b^2 - 4ac)]/ 2a not {b+sqrt(b^2-4ac)}/2a and {b-sqrt(b^2-4ac)}/2a. Also, 4ac is not 36 but 4*9*9 = 324.

To solve the equation 1/(x-1)+1/(x+1) = 18, we multiply both sides by the LCM of the denminators ,(x-1)(x+1) and we get:

x+1 +x-1 = 18(x-1)(x+1).

2x = 18(x^2-1).

2x = 18x^2-18/

We divide by 2:

x = 9x^2-9.

0 = 9x^2-x-9.

9x^2-x-9 = 0

We know the roots of ax^2+bx+c = 0 is given by:

x1 = {b+sqrt(b^2-4ac)}/2a and x2 = {b-sqrt(b^2-4ac)}/2a.

Here a = 9, b= -1 and c= -9.

Sothe roots of 9x^2-x-9 - 0 are:

x1 = {1+sqrt((-1)^2+4*9)}/2*9 = (1+sqrt37)/18..

and x2 = (1-sqrt37}/18