Will give brainliest to correct answer.

The Pyth. Thm. applies here:

(√x + 1)^2 + (2√x)^2 = (2√x + 1 )^2

Expanding the squares:

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

Let's subtract x + 2sqrt(x) + 1 + 4x from both sides:

4x + 4sqrt(x) + 1

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

3x + 2sqrt(x) - 4x = 0

Then 2sqrt(x) = x

Squaring both sides, 4x = x^2, or x^2 - 4x = 0. Then (x-4)x = 0, and the two possible solutions are 0 and 4.

Check these results by substitution. Does the Pyth. Thm. hold true for x=4?

idk lol xd

step-by-step explanation:

ez do it yourself lazy

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

you can download it here