How do you solve #sqrt(3x + 4) + x = 8# and find any extraneous solutions?

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Answer 1 · 2017-04-20T22:24:55

Please see the explanation.

Given: #sqrt(3x + 4) + x = 8#

Subtract x from both sides:

#sqrt(3x + 4) = 8-x#

Square both sides:

#3x + 4 = 64-16x+x^2#

Subtract 3x + 4 from both sides:

#0= x^2-19x+60#

This factors:

#(x - 4)(x - 15) = 0#

#x = 4 and x = 15#

Check:

#sqrt(3(4) + 4) + 4 = 8#
#sqrt(3(15) + 4) + 15 = 8#

#8 = 8#
#22=8#

#x = 4# is the solution.