Question

How do you solve `sqrt( 3x+1) - sqrt(x-1) = 2`?

Answer 1

`x = 1`

Explanation:

You're dealing with radical terms, which means that you need to make sure that the expressions that are unde the square root are always positive.

This means that you need

`3x + 1 >= 0 implies x >= -1/3`

`x - 1 >= 0 implies x >= 1`

Combining these two conditions will get you `x >= 1`. This means that any value of `x` that is smaller than `1` will cannot be a solution to the equation.

With this in mind, start by squaring both sides of the equation

`(sqrt(3x+1) - sqrt(x-1))^2 = 2^2`

`(sqrt(3x+1))^2 - 2sqrt((3x+1)(x-1)) - (sqrt(x-1))^2 = 4`

`3x + 1 - 2sqrt((3x+1)(x-1)) - x +1 = 4`

`2x - 2sqrt((3x+1)(x-1)) + 2 = 4`

Isolate the remaining square root on one side of the equation

`sqrt((3x+1)(x-1)) = (4-2-2x)/2 = 1 - x`

Once again, square both sides of the equation

`(sqrt((3x+1)(x-1)))^2 = (1-x)^2`

`(3x+1)(x-1) = 1 - 2x - x^2`

`3x^2 - color(red)(cancel(color(black)(2x))) - 1 = 1 - color(red)(cancel(color(black)(2x))) - x^2`

Rearrange the quadratic equation by moving all the terms on one side of the equation

`4x^2 -2 = 0`

This is equivalent to

`2(x^2-1) = 0 implies x^2 - 1 = 0 <=> {(x=-1), (x=1) :}`

Since you need `x` to satisfy the condition `x>=1`, only `x=1` will be a valid solution; `x = -1` will be an extraneous solution.

Do a quick check to make sure that the calculations are correct

`sqrt(3 * (1) +1) - sqrt((1)-1) = 2`

`sqrt(4) - 0 = 2`

`2 = 2color(white)(x)color(green)(sqrt())`