# If (-2 + x^2)^5 = 1 then what can x be equal to in the set of positive irrational numbers?

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( -2 + x^2 )^5 = 1

We need to determine x value which is an irrational number :

First we will raise to the (1/5) power for both sides:

==>[ (-2 + x^2 )^5]^1/5 = (1^1/5)

==> (-2 + x^2 ) = 1

Now add 3 to both sides:

==> x^2 = 3

Now we will take the square root for both sides:

==> x1 = sqrt3

==> x2= - sqrt3

Then there are two possible solutions for x which they are:

**x = { sqrt3 , - sqrt3}**

To find the positive irrational solutions of (-2+x)^5 = 1.

We take the (-2+x^2)^5 = 1.

Put -2+x^2 = y.

Then y ^5 = 1.

Then y^5-1 = 0

(y^5-1) = (y-1)(y^4+y^3+y^2+y+1).

Therefore y = 1 is the only real solutions, as all other solutions are imaginary.

Therefore y = 1.

Now -2+x^2 = 1.

Therefore x^2 = 1+2.

x = +sqrt3, or x= -sqrt3.

so only +sqrt3 is the positive irrational solution of (-2+x^2)^5 = 1.

Therefore set of positive irrational solutions = {sqrt3}.