Question

How do you solve `1/((x - 1)(x - 2)) + 1/((x - 2)(x - 3) )+ 1/((x - 3)(x - 4)) = 1/6`?

Answer 1

`x=-2` or `x=7`

Explanation:

Given:

`1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4)) = 1/6`

Multiply through by `6(x-1)(x-2)(x-3)(x-4)` to get:

`6(x-3)(x-4)+6(x-1)(x-4)+6(x-1)(x-2) = (x-1)(x-2)(x-3)(x-4)`

Multiply out:

`6(x^2-7x+12)+6(x^2-5x+4)+6(x^2-3x+2) = x^4-10x^3+35x^2-50x+24`

`rarr`

`18x^2-90x+108 = x^4-10x^3+35x^2-50x+24`

Subtract the left hand side from the right to get:

`x^4-10x^3+17x^2+40x-84 = 0`

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-84` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1`, `+-2`, `+-3`, `+-4`, `+-6`, `+-7`, `+-12`, `+-14`, `+-21`, `+-28`, `+-42`, `+-84`

Substituting `x=2` into the quartic we find:

`x^4-10x^3+17x^2+40x-84 = 16-80+68+80-84 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^4-10x^3+17x^2+40x-84 = (x-2)(x^3-8x^2+x+42)`

Substituting `-2` into this cubic we find:

`x^3-8x^2+x+42 = -8-32-2+42 = 0`

So `x=-2` is a zero and `(x+2)` a factor:

`x^3-8x^2+x+42 = (x+2)(x^2-10x+21)`

To factor and find the zeros of the remaining quadratic, note that `3+7 = 10` and `3*7 = 21`, so:

`x^2-10x+21 = (x-3)(x-7)`

So the remaining zeros are `x=3` and `x=7`.

So all the zeros of our quartic polynomial are:

`-2, 2, 3, 7`

Note that the values `2` and `3` are not solutions of the original rational equation, since they result in zero denominators.

So the solutions of the original rational equation are `x=-2` and `x=7`