Answer: x = 0 and x = 16.
Explanation:
1) given
[tex] \sqrt{2x+4} - \sqrt{x} = 2[/tex]
2) transpose √x to the right side
[tex] \sqrt{2x+4} = \sqrt{x} +2[/tex]
3) square both sides
2x + 4 = (√x + 2)²
2x + 4 = x + 4√x + 4
4) transpose 4√x + 4 to the left side
2x - x + 4 - 4 = 4√x
x = 4√x
5) square both sides
x² = 16x
6) transpose 16x to the left side
x² - 16x = 0
7) factor
x (x - 16) = 0
8) equal each factor to 0
x = 0
x - 16 = 0 ⇒ x = 16.
9) test both solutions to discard extraneous solutions:
a) x = 0
√(2(0) + 4) - √0 = 2
√4 = 2
2 = 2 ⇒ ok
b) x = 16
√(2(16) + 4) - √16 = 2
√(32 + 4) - 4 = 2
√36 - 4 = 2
6 - 4 = 2
2 = 2 ⇒ ok.
So, both x = 0 and x = 16 are solutions.
