# How would solve equation (`sqrt(5+2sqrt6)` )^x+(`sqrt(5-2sqrt6)` )^x=98?

You should notice that you may convert both radicands into two squares, such that:

`5 + 2sqrt6 = (sqrt3)^2 + 2*sqrt3*sqrt2 + (sqrt2)^2 = (sqrt3 + sqrt2)^2`

`5 - 2sqrt6 = (sqrt3)^2 - 2*sqrt3*sqrt2 + (sqrt2)^2 = (sqrt3 - sqrt2)^2`

Replacing `(sqrt3 + sqrt2)^2` for `5 + 2sqrt6` and `(sqrt3 - sqrt2)^2 ` for `5 - 2sqrt6` yields:

`(sqrt((sqrt3 + sqrt2)^2))^x + (sqrt((sqrt3 - sqrt2)^2))^x = 98`

Since `sqrt(a^2) = |a|` yields:

`|sqrt3 + sqrt2|^x + |sqrt3 - sqrt2|^x = 98`

`(sqrt3 + sqrt2)^x + (sqrt3 - sqrt2)^x = 98`

You need to notice that `sqrt 3 + sqrt 2 = 1/(sqrt 3 - sqrt 2),` hence, raising to `x` both sides, yields:

`(sqrt3 + sqrt2)^x = 1/(sqrt 3 - sqrt 2)^x`

You need to come up with the following substitution, such that:

`(sqrt3 + sqrt2)^x = y => (sqrt 3 - sqrt 2)^x = 1/y`

Replacing ` y` for `(sqrt3 + sqrt2)^x` and `1/y` for `(sqrt 3 - sqrt 2)^x` , yields:

`y + 1/y = 98 => y^2 - 98y + 1 = 0`

You need to use quadratic formula, such that:

`y_(1,2) = (98+-sqrt(98^2 - 4))/2`

`y_(1,2) = (98+-40sqrt6)/2 => y_(1,2) = 49 +- 20sqrt6`

You need to solve for x the following equations, such that:

`(sqrt3 + sqrt2)^x = 49 +- 20sqrt6`

Taking common logarithms both sides, yields:

`ln (sqrt3 + sqrt2)^x = ln(49 +- 20sqrt6)`

Using the power property of logarithms, yields:

`x*ln (sqrt3 + sqrt2) = ln(49 +- 20sqrt6) => x_(1,2) = ln(49 +- 20sqrt6)/ln (sqrt3 + sqrt2)`

Hence, evaluating the solutions, to the given equation, yields `x_(1,2) = ln(49 +- 20sqrt6)/ln (sqrt3 + sqrt2).`

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