Question

How do you solve `4-sqrt(x-3)=sqrt(x+5)`?

Answer 1

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

Square both sides giving
`16 - 8sqrt(x-3) + (x-3) = x+5`

`13 - 8sqrt(x-3) + cancel(x) = cancel(x) + 5`

`-8sqrt(x-3) = -8`

`sqrt(x-3) = 1`

Square both sides again
`x-3 = 1`

`x = 4`

Verify by substituting `x=4` back into the original equation.

Answer 2

`x=4`

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

Squaring both sides

`[4-sqrt(x-3)]^2 = [sqrt(x+5)]^2`

As `(a-b)^2 = a^2-2ab+b^2`

So,

`4^2-8sqrt(x-3)+[sqrt(x-3)]^2 = [sqrt(x+5)]^2`

`16-8sqrt(x-3)+(x-3) = x+5`

`16+x-3-x-5-8sqrt(x-3)=0`

`16-8-8sqrt(x-3) =0`

`8-8sqrt(x-3)=0`

`8[1-sqrt(x-3)] =0`

`1-sqrt(x-3)=0`

`sqrt(x-3) =1`

Again Squaring both sides

`[sqrt(x-3)]^2 =1 ^2`

`x-3 =1`

`x =1+3`

`x=4`