Question

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

Answer 1

`{4}`

Explanation:

We start by squaring both sides.

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

`x + 5 = 25 - 10sqrt(x) + x`

`10sqrt(x) = 20`

`sqrt(x) = 20/10`

`sqrt(x) = 2`

We square both sides one more time.

`x = 4`

Now we check our answer.

`sqrt(4 + 5) = 5 - sqrt(4)`

`3 = 5 - 2 color(green)(√)`

Practice exercises

`1`. Solve for `x`. Make sure to verify your solutions because they may or not be extraneous.

a) `sqrt(2x - 2) = sqrt(x) + 1`
b) `sqrt(3x - 5) = sqrt(x + 6) + 1`

Solutions

`1`. a) `x = 9`
b) `x = 10`

Hopefully this helps, and good luck!

Answer 2

I got `x=4`

Explanation:

We can try write it as:
`sqrt(x+5)+sqrt(x)=5`
square both sides:
`(sqrt(x+5)+sqrt(x))^2=5^2`
`x+5+2sqrt(x+5)sqrt(x)+x=25`
`2sqrt(x+5)sqrt(x)=20-2x`
square again:
`4x(x+5)=400-80x+4x^2`
`cancel(4x^2)+20x=400-80x+cancel(4x^2)`
`100x=400`
`x=400/100=4`