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

Question

2022 views around the world

You can reuse this answer
Creative Commons License

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

$18x - 15x = 9$

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

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