If (sqrt{x} + sqrt{x + 1} = sqrt{2x + 1}), find the possible values of x.
-
A.
1 and -1 -
B.
-1 and 2 -
C.
1 and 2 -
D.
0 and -1
Correct Answer: Option 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 + 2sqrt{x(x + 1)} + x + 1 = 2x + 1)
(2x + 1 + 2sqrt{x(x+1)} – (2x + 1) = 0)
((2sqrt{x(x + 1)})^{2}= 0^{2} implies 4(x(x + 1)) = 0)
(therefore x(x + 1) = 0)
(x = text{0 or -1})