How do you solve #sqrt( 5x+9) + 3 = 2x#?

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Answer 1 · 2016-08-03T02:06:00

#sqrt(5x + 9) = 2x - 3#

#(sqrt(5x + 9))^2 = (2x - 3)^2#

#5x + 9 = 4x^2 - 12x + 9#

#0 = 4x^2 - 17x#

#0 = x(4x - 17)#

#x = 0 and 17/4#

Checking in the original equation we find the #x = 0# is extraneous. Hence, the solution set is #{17/4}#.

Hopefully this helps!