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x=2+2+2+2+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√x=2+2+2+2+... --> x=2+(2+2+2+...−−−−−√)−−−−−−−−−−√−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−√x=2+(2+2+2+...), as the expression under square root extends infinitely then expression in brackets would equal to xx itself so we can rewrite given expression as x=2+x−−−−−√x=2+x. Square both sides x2=2+xx2=2+x --> x=2x=2 or x=−1x=−1. As given that x>1x>1 then only one solution is valid: x=2x=2.
Answer: B.
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