How do you solve #sqrtx -sqrt(x-1) = 1?

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Answer 1 · 2016-05-18T03:26:23

#sqrt(x) - sqrt(x - 1) = 1#

#sqrt(x) = 1 + sqrt(x - 1)#

#(sqrt(x))^2 = (1 + sqrt(x - 1))^2#

#x = 1 + x - 1 + 2sqrt(x - 1)#

#0 = 2sqrt(x - 1)#

#(0)^2 = (2sqrt(x - 1))^2#

#0 = 4(x - 1)#

#0 = 4x - 4#

#4 = 4x#

#x = 1#

Therefore, #{x = 1}#. Checking the solution back in the original equation we find it works.

Hopefully this helps!