How do you solve `(sqrt(x + 7)) - 2(sqrt(x)) = -2` and find any extraneous solutions?

The first thing to do here is write out the conditions that a value of `x` must satisfy in order to be a valid solution to the equation

`sqrt(x+7) - 2sqrt(x) = -2`

You're working with real numbers, so right from the start you know that

• you can only take the square root of a positive number
• the square root of a positive number will always return a positive number

In your case, this means that you must have

`x + 7 >= 0 implies x >= -7`

`x >= 0`

Combine these two conditions to get

`x in [-7, + oo) nn [0, + oo) implies color(purple)(|bar(ul(color(white)(a/a)color(black)(x in [0, + oo))color(white)(a/a)|)))`

Now, your goal here is to get rid of the radical terms. To do that, square both sides of the equation

`(sqrt(x+7) - 2sqrt(x))^2 = (-2)^2`

This will get you

`x + 7 - 4 * sqrt((x+7) * x) + 4x = 4`

Isolate the remaining radical term on one side of the equation

`-4 * sqrt( (x+7) * x) = -5x - 3`

`4 * sqrt((x+7) + x) = 5x + 3`

Square both sides of the equation again

`(4 * sqrt((x+7) * x))^2 = (5x + 3)^2`

`16 * (x^2 + 7x) = 25x^2 + 30x + 9`

Rearrange to quadratic form

`9x^2 - 82x + 9 = 0`

Use the quadratic formula to find the two roots

`x_(1,2) = (-(82) +- sqrt( (-82)^2 - 4 * 9 * 9))/(2 * 9)`

`x_(1,2) = (82 +- sqrt(6400))/18`

`x_(1,2) = (82 +- 80)/18 implies {(x_1 = (82 + 80)/18 = 9), (x_2 = (82-80)/18 = 1/9) :}`

Since you have

`x_1, x_2 in [0, +oo)`

both roots can be valid solutions to the original equation. Do a quick calculation to see if both roots are indeed valid solutions

`sqrt(9 + 7) - 2 * sqrt(9) = -2`

`4 - 2 * 3 = -2 " "color(green)(sqrt())`

`sqrt(1/9 + 7) - 2 * sqrt(1/9) = -2`

`sqrt(64/9) - 2 * 1/3 = -2`

`8/3 - 2 * 1/3 = -2 " "color(red)(xx)`

As you can see, `x = 1/9` is not a solution to the original equation, i.e. it's an extraneous solution. Therefore, you can say that the original equation has one valid solution, `x=9`.