Find the solution of sqrt x+2+4=6, and determine if it is an extraneous solution.

Part (1): getting the value of x:
To find the value of x, we would need to isolate the x on one side of the equation as follows:
[tex] \sqrt{x+2} + 4 = 6 [/tex]
[tex] \sqrt{x+2} = 6 - 4 = 2[/tex]
[tex]( \sqrt{x+2} )^{2} = (2)^2[/tex]
[tex]x + 2 = 4[/tex]
[tex]x = 4 - 2 [/tex]
[tex]x = 2[/tex]

Part (2): determining if it is an extraneous solution:
An extraneous solution is defined as a solution of the simplified version of the equation that does not satisfy the original equation.
Therefore, to determine whether the solution you got is extraneous or not, substitute with it in the original equation and check if it satisfies the equation or not.
From part 1, we calculated that x = 2.
Substitute with x = 2 in the original equation as follows:
[tex] \sqrt{x+2} + 4 = \sqrt{2+2} + 4 = \sqrt{4} + 4 = 2 + 4 = 6[/tex]
The solution satisfies the equation. Therefore, it is not an extraneous solution.

Hope this helps :)