2009/023) Find values of x: √x + √(x+2) + √[x(x+2)] = 3-x

as product of 1st term and 2nd term is 3rd I would take the 3rd term to the right and get

√x + √(x+2) = (3-x)- √[x(x+2)]

square both sides to get
x + (x+2) + 2√[x(x+2)] = (3-x)^2 + x(x+2)- 2(3-x) √[x(x+2)]

or
2x + 2 + 2√[x(x+2)] = 9-6x+ x^2 + x^2 + 2x - 2(3-x) √[x(x+2)]
= 9-4x+ 2x^2 +(2x-6) √[x(x+2)]

now keeping redicals on one side we get
2x + 2 - (2x^2-4x+9) = (2x-8) √[x(x+2)]
(2x-8) √[x(x+2)] = -2x^2 +6x -7
now by squaring again you can get rid of redicals and proceed

No one has solved so far so I continue

(2x-8)^2x(x+2) = (-2x^2+6x-7)^2

or expanding we get after simplfication

64x^2-212x + 49 = 0

or(4x-1)(16x-49) = 0

x = 1/4 or 49/16
x has to be less than 3 as LHS > 0

so x = 1/4 need to be checked and this is indeed a solution

so x = 1/4