Question

How do you solve `sqrt(x+4) = sqrt(x) + sqrt(2)`?

Answer 1

`x = 1/2`

Explanation:

Right from the start, you know that, for real numbers, you can't take the square root of a negative value, which means that you need to have

`x >=0`

Next, square both sides of the equation to get

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

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

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

This is equivalent to

`sqrt(2x) = 1`

Once again, you have

`(sqrt(2x))^2 = 1^2`

`2x = 1 implies x = 1/2`

Since this value of `x` satisfies the condition `x >= 0`, it will be a valid solution to the original equation.

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

`sqrt(1/2 + 4) = sqrt(1/2) + sqrt(2)`

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

`3/sqrt(2) = 3/sqrt(2)color(white)(x)color(green)(sqrt())`