How do you solve #sqrt(2x +3) = 6-x#?

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Answer 1 · 2018-04-04T21:31:19

#x = 3#

#sqrt (2x + 3) = 6 - x#

Square both sides:

#sqrt (2x + 3)^2 = (6 - x)^2#

Notice that #2x + 3 >= 0 # and #6 - x >= 0#
#=> -3/2 <= x <= 6 #

#2x + 3 = 36 - 12x + x^2 #

#x^2 - 14x + 33 = 0#

#(x - 11)(x - 3) = 0#

#x = 3, 11#

Since #-3/2 <= x <= 6 # , #x = 11# will not work in the original eqaution and the answer is #x = 3#.