If \(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\), find the possible values of
FURTHER MATHEMATICS
WAEC 2013
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: D. 0 and -1
Explanation
\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)
Squaring both sides, we have
\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)
\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)
\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)
\((2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0\)
\(\therefore x(x + 1) = 0\)
\(x = \text{0 or -1}\)
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