solve equation for X: sqrt(x-2) = sqrt(3) +sqrt(x)?

Square both sides, you get x-2 = 3 + 2sqrt(3x) + x.

That simplifies to 2sqrt(3x) = -5

Square both sides: 12x = 25

So x = 25/12.

But if you substitute into the original equation, sqrt(x-2) = sqrt(1/12) = sqrt(3)/6

While sqrt(3)+sqrt(x) = sqrt(3)+sqrt(25/12) = sqrt(3) + 5 sqrt(3)/6

Thus 25/12 is a superfluous root, and in fact there are no solutions.

This can also be seen by graphing y=sqrt(x-2) and y=sqrt(3)+sqrt(x). One is the square root graph shifted 2 units to the right, the other is the square root graph shifted sqrt(3) units upward. These two graphs do not intersect.

sqrt(x-2) = sqrt(3) +sqrt(x)

sqrt(x-2) - sqrt(3) = sqrt(x)

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

x+1 - 2*sqrt(3x-6) = x

1 = 2*sqrt(3x-6)

1= 4*(3x-6)

25=12x

x=25/12

√(x-2) = √3 + √x

Square both sides

(√(x-2))² = (√3 + √x)²

x - 2 = 3 + 2√3√x + x

subtract 3 and subtract x from both sides:

x - 2 - 3 - x = 3 + 2√3√x + x - 3 - x

-5 = 2√3√x

square both sides

25 = 4(3)x

25/12 = x

square both sides to remove the squared...

that will give you x-2=(sqrt(3)+sqrt(x))(sqrt(3)+sqrt(x))

factorize x-2=(3+sqrt(3)sqrt(x)+sqrt(3)sqrt(x)+x)

x-2-3-x=2sqrt3sqrt(x)

-6/2=sqrt(3)sqrt(x)

-3=sqrt(3)sqrt(x)

-3/sqrt(3)=sqrt(x)

sqrt(3)=sqrt(x)

square both sides

3=x

x=3