How do you solve #sqrt(2x+3)-sqrt(x+2)=2# and find any extraneous solutions?

Question

5210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-19T15:55:26

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

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

#2x + 3 = 4 + 4sqrt(x + 2) + x+ 2#

#2x + 3 - 4 - x - 2 = 4sqrt(x + 2)#

#x - 3 = 4sqrt(x + 2)#

#(x - 3)^2 = (4sqrt(x + 2))^2#

#x^2 - 6x + 9 = 16(x + 2)#

#x^2 - 6x + 9 = 16x + 32#

#x^2 - 22x -23 = 0#

#(x - 23)(x + 1) = 0#

#x = 23 and x = -1#

Checking in our original equation, we find that #x = 23# works but #x = -1# does not.

Our solution set is therefore #{x = 23}#.

Hopefully this helps!