# what is equationn squarerot(x+2-2squareroot(x+1))+squareroot(x+2+2squareroot(x+1))-4=0?

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You need to solve for x the equation `sqrt(x+2-2sqrt(x+1))+sqrt(x+2+2sqrt(x+1))-4=0` .

You should keep to the left side only the square roots such that:

`sqrt(x+2-2sqrt(x+1))+sqrt(x+2+2sqrt(x+1)) = 4`

Raising to square both sides yields:

`x + 2 - 2sqrt(x+1) + 2 sqrt((x+2)^2 - 4(x+1)) + x + 2 +2sqrt(x+1) = 4`

Reducing like terms yields:

`2x + 4 + 2 sqrt((x+2)^2 - 4(x+1)) = 4` `2x + 2 sqrt((x+2)^2 - 4(x+1)) = 0`

Subtracting 2x both sides yields:

`2 sqrt((x+2)^2 - 4(x+1` `))` = -2

Reducing by 2 yields:

`sqrt((x+2)^2 - 4(x+1)) = -1`

Notice that there is no real value of x to yield -1 if it is squared.

I will prove to you that this statement in the next steps.

Raising to square yields:

`(x+2)^2 - 4(x+1)) = 1`

Expanding the binomial and opening brackets yields:

`x^2 + 4x + 4 - 4x - 4 = 1`

`` You need to reduce like terms such that:

`x^2 = 1 =gt x_(1,2) = +-1`

Plugging x = 1 in the equation yields:

`sqrt(1+2-2sqrt(1+1))+sqrt(1+2+2sqrt(1+1))-4=0`

`sqrt(3-2sqrt(2))+sqrt(3+2sqrt(2)) = 4`

`3 - 2sqrt2 + 2sqrt(9 - 8) + 3 + 2sqrt2 = 16`

6 + 2 = 16 contradiction

Plugging x = -1 in the equation yields:

`sqrt(1-2sqrt(0))+sqrt(1+2sqrt(0)) = 4`

1+1 = 4 contradiction

**Hence, there is no real solution to this equation.**