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)
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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?
