How do you solve #sqrt(3x-5)-sqrt(3x)=-1#?

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Answer 1 · 2015-08-05T13:13:59

#x = 3#

The first thing you need to do is square both sides of the equation. This will leave you with only one radical term.

#(sqrt(3x-5) - sqrt(3x))^2 = (-1)^2#

#[(sqrt(3x-5))^2 - 2 * sqrt( (3x-5) * 3x) + (sqrt(3x))^2] = 1#

#(3x - 5 - 2sqrt(9x^2 - 15x) + 3x) = 1#

#6x - 5 - 2sqrt(9x^2-15x) = 1#

Isolate the remaining radical term on onse side of the equation, then square both sides again.

#-2sqrt(9x^2 - 15x) = 6 -6x#

This is equivalent to

#sqrt(9x^2 - 15x) = 3x - 3#

#(sqrt(9x^2 - 15x))^2 = (3x-3)^2#

#color(red)(cancel(color(black)(9x^2))) - 15x = color(red)(cancel(color(black)(9x^2))) - 18x + 9#

Rearrange to isolate #x# on onse side of the equation

#18x - 15x = 9#

#3x = 9 => x = 9/3 = color(green)(3)#