Question

How do you solve `sqrt (x) + sqrt(x-5) = 5`?

Answer 1

`x = 9`

Explanation:

The first thing to do is figure out which values of `x` cannot be a solution to the equation.

For real numbers, you need to take the square root of positive numbers only, which means that you will need

`x - 5 >= 0 implies x >= 5`

Next, square both sides of the equation to reduce the number of radical terms from two to one.

`(sqrt(x) + sqrt(x-5))^2 = 5^2`

`(sqrt(x))^2 + 2sqrt(x(x-5)) + (sqrt(x-5))^2 = 25`

`x + 2sqrt(x(x-5)) + x - 5 = 25`

Rearrange to get the radical term alone on one side of the equation

`2sqrt(x(x-5)) = 30 - 2x`

`sqrt(x(x-5)) = 15 - x`

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

`(sqrt(x(x-5)))^2 = (15 - x)^2`

`x(x-5) = 225 - 30x + x^2`

`color(red)(cancel(color(black)(x^2))) - 5x = 225 - 30x + color(red)(cancel(color(black)(x^2)))`

`25x = 225 implies x = color(green)(9)`

Since `x = 9` satisfies the condition `x>= 5`, this will be a valid solution to the original equation.

Do a quick check to make sure that the calculations came out right

`sqrt(9) + sqrt(9-5) = 5`

`3 + 2 = 5color(white)(x)color(green)(sqrt())`