# alg. 2 nth roots topicstress on applications with radicals in different order

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You need to know that if an equation contains two or more radicals having different orders, either you will bring the radicals to a common order, or, you will make a change of variable, that will make the process of evaluation less complicated.

Considering as example the following equation, you will notice how to perform the change of the variable to simplify the evaluation.

`root(3)(9-x) + sqrt(x-1) = 2`

Notice that the radicals have different orders, hence, you may change the variable such that:

`x - 1 = t => x = t + 1`

`9 - x = 9 - t - 1 = 8 - t`

`roo(3)(8 - t) + sqrt t = 2`

You need to isolate `roo(3)(8 - t)` to the left, such that:

`roo(3)(8 - t) = 2 - sqrt t`

You need to raise to cube both sides, such that:

`8 - t = 8 - (sqrt t)^3 - 6sqrt t(2 - sqrt t)`

Reducing duplicate terms yields:

`-t = - (sqrt t)^3 - 6sqrt t(2 - sqrt t)`

`t = (sqrt t)^3 + 6sqrt t(2 - sqrt t)`

`t - (sqrt t)^3 - 6sqrt t(2 - sqrt t) = 0`

Factoring out `sqrt t` yields:

`sqrt t(sqrt t - t - 12 + 6sqrt t) = 0`

`sqrt t = 0 => t = 0`

`7sqrt t - t + 12 = 0`

`-t + 7sqrt t + 12 = 0 => t - 7sqrt t - 12 = 0`

You may perform the next substitution such that:

`sqrt t = y => t = y^2`

`y^2 - 7y - 12 = 0`

Using quadratic formula yields:

`y_(1,2) = (7+-sqrt(49 + 48))/2 => y_(1,2) = (7+-sqrt97)/2`

Since `sqrt t = y => sqrt t = (7+sqrt97)/2 => t = (7+sqrt97)^2/4`

`x = t+1 => x = (7+sqrt97)^2/4 + 1`

**Hence, using the indicated substitutions, as in the equation above, you may make less complicated the process of evaluation of equation.**