# equation with rsdicals and substituents examplesgive 2 examples with substituent in radicals eqs

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You may consider as an example the following equation that contains radicals of different orders, such that:

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

You sould come up with the following substitution, such that:

`x + 2 = t => x + 3 = t + 1`

Changing the variable yields:

`sqrt(t + 1) - root(3) t = 1`

You need to isolate `sqrt(t + 1)` to the left such that:

`sqrt(t + 1) = 1 + root(3) t`

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

`t + 1 = 1 + 2root(3)t + root(3)(t^2)`

Reducing duplicate terms yields:

`t -root(3)(t^2) = 2root(3)t`

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

`t^3 - t^2 - 3t root(3)(t^2)(t - root(3)(t^2)) = 8t`

Since `t -root(3)(t^2) = 2root(3)t` , you may substitute `2root(3)t` for `t -root(3)(t^2)` , such that:

`t^3 - t^2 - 3t root(3)(t^2)(2root(3)t) = 8t`

`t^3 - t^2 - 6t^2 - 8t = 0`

`t^3 - 7t^2 - 8t = 0`

Factoring out t yields:

`t(t^2 - 7t - 8) = 0 => t = 0`

`t^2 - 7t - 8 = 0 => t_(1,2) = (7+-sqrt(49 + 32))/2`

`t_(1,2) = (7+-sqrt81)/2 => t_(1,2) = (7+-9)/2 => t_1 = 8 ; t_2 = -1`

Substituting back `x + 2` for t yields:

`x + 2 = 0 => x = -2`

`x + 2 = 8 => x = 6`

`x + 2 = -1 => x = -3`

**Hence, performing the indicated substitutions, you may solve an equation that contains radicals of different orders.**