Question

How do you solve and check for extraneous solutions in `3sqrt(2x+4)=12`?

Answer 1

`x = 6`

Explanation:

First, isolate the radical on one side of the equation by dividing both sides by ##

`(color(red)(cancel(color(black)(3))) * sqrt(2x+4))/color(red)(cancel(color(black)(3))) = 12/3`

`sqrt(2x+4) = 4`

Square both sides of the equation to get rid of the square root

`(sqrt(2x+4))^2= 4^2`

`2x+4 = 16`

Finally, isolate `x` on one side of the equation by adding `-4` to both sides and dividing everything by `2`

`2x + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = 16 - 4`

`(color(red)(cancel(color(black)(2)))x)/color(red)(cancel(color(black)(2))) = 12/2`

`x = color(green)(6)`