Question

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

Answer 1

x = 4.

Explanation:

I have deleted my inclusion of x=0 as a solution. Thanks to Veddesh, for duly pointing out the mistake..
`x>=0`.
`(sqrt(5+x))^2=(5-sqrtx)^2`
This simplifies to `sqrtx=2`. So, x = 4.

Answer 2

`x=4`

Explanation:

`color(blue)(sqrt(5+x)+sqrtx=5`

Subtract `sqrtx` both sides

`rarrsqrt(5+x)+cancel(sqrtx-sqrtx)=5-sqrtx`

`rarrsqrt(5+x)=5-sqrtx`

Square both sides to get rid of the radical sign

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

In the right hand side use the formula

`color(brown)((a-b)^2=a^2-2ab+b^2`

So,

`rarr5+x=25-10sqrtx+x`

Subtract `x` and `5` both sides

`rarrcancel(5+x-5-x)=25-10sqrtx+x-5-x`

`rarr0=20-10sqrtx`

Rewrite the equation in standard form

`rarr20-10sqrtx=0`

Subtract `20` both sides

`rarr20-10sqrtx-20=-20`

`rarr-10sqrtx=-20`

Divide both sides by `-10`

`rarr(cancel(-10)sqrtx)/cancel(-10)=(-20)/-10`

`rarrsqrtx=2`

Square both sides

`rarrx=2^2`

`color(green)(rArrx=4`