If \(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\), find the possible values of x.
- 1 and -1
- -1 and 2
- 1 and 2
-
0 and -1
The correct answer is: D
Explanation
\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)
Squaring both sides, we have
\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)
\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)
\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)
\((2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0\)
\(\therefore x(x + 1) = 0\)
\(x = \text{0 or -1}\)