How do you solve and check for extraneous solutions in `sqrt(3x - 2) + 3 =4x`?

Start by isolating the radical term on one side of the equation. This can be done by adding `-3` to both sides

`sqrt(3x-2) + color(red)(cancel(color(black)(3))) - color(red)(cancel(color(black)(3))) = 4x - 3`

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

Now, because you can't take the squaree root of a negative number if you're working with real numbers, any solution you will find must satisfy two conditions

`sqrt( underbrace((3x-2))_(color(orange)("must be positive"))) = overbrace(4x-3)^(color(red)("must be positive"))`

• `3x-2>=0 implies x >=2/3`
• `4x-3 >=0 implies x >=3/4`

Overall, you need `x>= 3/4`.

Square both sides of the equation to remove the radical term

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

`3x-2 = 16x^2 - 24x + 9`

Rerarrange to get

`16x^2 - 27x + 11 = 0`

Use the quadratic formula to find the two solutions to this quadratic equation

`x_(1,2) = (-(-27) +- sqrt( (-27)^2 - 4 * 16 * 11))/(2 * 16)`

`x_(1,2) = (27 +- sqrt(25))/32`

`x_(1,2) = (27 +- 5)/32 = {( x_1 = (27 + 5)/32 = 1), (x_2 = (27-5)/32 = 11/16) :}`

Notice that only `x_1` satisfies the condition `x>=3/4`, which means that `x_2 = 11/16` will be an extraneous solution.

This is of course because

`11/16 <12/16 = 3/4`

Therefore, your original equation will only have one solution, `x = color(green)(1)`.

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

`sqrt(3 * 1 -2) = 4 * (1)-3`

`sqrt(1) = 1 <=> 1 = 1 color(green)(sqrt())`