Question

How do you solve `sqrt(x+1) + 1=sqrt (2x)`?

Answer 1

`x=8`

Explanation:

`sqrt(x+1)+1=sqrt(2x), x>0`

`1=sqrt(x+1)-sqrt(2x)` subtract `sqrt(x-1)` from both sides

`1=2x-2sqrt(2x(x+1))+(x+1)` square both sides

`0=3x-2sqrt(2x(x+1))` collect terms and subtract `1` from both sides

`sqrt(2x(x+1))=(3x)/2` add `sqrt(2x(x+1))` to both sides

`2x(x+1)=(9x^2)/4` square both sides

`2x^2+2x=(9x^2)/4` expand brackets

`0=x^2/4-2x` move terms to RHS

`0=x(x/4-2)` factorise

`x=0 or x=8` solve linear equations

Since `x>0`, `x=8`

Answer 2

`x=8`

Explanation:

Given:

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

Square both sides (noting that squaring can introduce extraneous solutions) to get:

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

Subtract `x+2` from both sides to get:

`2sqrt(x+1) = x-2`

Square both sides to get:

`4(x+1) = x^2-4x+4`

Subtract `4x+4` from both sides to get:

`0 = x^2-8x = x(x-8)`

So `x=0` or `x=8`

Check whether these are solutions of the original equation:

`sqrt((color(blue)(0))+1)+1 = 2 != 0 = sqrt(2(color(blue)(0)))`

`sqrt((color(blue)(8))+1)+1 = 3+1 = 4 = sqrt(16) = sqrt(2(color(blue)(8)))`

So `x=8` is the only solution of the given equation.