Question

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

Answer 1

`x = 3`

Explanation:

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)`