Question

3x+1/x^(2)-1=2/x-2+x/x-1

Answer 1

x = 0 and 3 other real roots. See below.

Explanation:

This is what you wrote:

`3x + (1)/(x^2) - 1 = (2)/(x) - 2 + (x)/(x) - 1`

I don't think that is what you meant to write.

I think you meant to write this:

`3x + (1)/(x^2 - 1) = (2)/(x - 2) + (x)/(x - 1)`

I'm going to answer the second equation because that is what I think you meant.

`3x + (1)/((x - 1)(x + 1)) = (2)/(x - 2) + (x)/(x - 1)`

`3x(x - 1)(x + 1)(x - 2) + (x - 2) = 2(x - 1)(x + 1) + x(x + 1)(x - 2)`

`(3x^2 - 6x)(x^2 - 1) + x - 2 = 2x^2 - 2 + x(x^2 - x - 2)`

`3x^4 - 3x^2 - 6x^3 + 6x<strong> </strong>+ x <strong>- 2</strong> = 2x^2 <strong>- 2</strong> + x^3 - x^2 <strong>- 2x</strong>`

`3x^4 - 7x^3 - 4x^2 + 9x = 0`

`x(3x^3 - 7x^2 - 4x + 9) = 0`

`f(x) = 3x^3 - 7x^2 - 4x + 9`

`f(-3) = -123`

`f(-2) = -35`

`f(-1) = 3`

`f(0) = 9`

`f(1) = 1`

`f(2) = -3`

`f(3) = 15`

One root is x = 0.

There is a root between x = -2 and x = -1.

There is a root between x = 1 and x = 2.

There is a root between x = 2 and x = 3.

Plot the graph of f(x) = 3x^3 - 7x^2 - 4x + 9 and try to read the other three roots.