How do you solve #x=sqrt( -7x+8)# and find any extraneous solutions?

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Answer 1 · 2016-07-16T04:05:44

#(x)^2 = (sqrt(-7x + 8))^2#

#x^2 = -7x + 8#

#x^2 + 7x - 8 = 0#

#(x + 8)(x - 1) = 0#

#x = -8 and 1#

Checking in the original équation, we find only #x = 1# works. Hence, #{1}# is the solution set.

Hopefully this helps!