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

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Answer 1 · 2015-10-14T00:52:51

Move one of the radicals to the other side of the equation, then square both sides a couple of times. Answer: #x=14#

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

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

#x-5 = 2x+1-4sqrt(2x-3)#

#-x-6= -4sqrt(2x-3)#

#(-x-6)^2= (-4sqrt(2x-3))^2#

#x^2+12x+36 = 32x-48#

#x^2-20x+84=0#

#(x-6)(x-14)=0#

#x=6 or x=14#

Check for extraneous solutions:

Only x = 14 works in the original equation:

#sqrt(14-5) = sqrt(2*14-3)-2#

#3 = 3#

Hope that helped