How do you solve #sqrt(10–x)+x=8# and find any extraneous solutions?

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Answer 1 · 2016-09-26T02:29:38

#sqrt(10 - x) = 8 - x#

#(sqrt(10 - x))^2 = (8 - x)^2#

#10 - x = 64 - 16x + x^2#

#0 = x^2 - 15x + 54#

#0 = (x - 9)(x - 6)#

#x = 9 and 6#

Checking in the original equation, you will find that #x = 9# is extraneous. However, #x = 6# works. Hence, our solution set will be #{6}#.

Hopefully this helps!