The radical sum

$\sqrt {x+ 2} + \sqrt {x} = 2$

$\sqrt {x+ 2} = 2 - \sqrt {x}$

After squaring both sides, we get:

$x+ 2 = 4 - 4 \sqrt {x} + x$

$4 \sqrt {x} = 2$

$\sqrt {x} = 0.5$

$x = \boxed {0.25}$

$\sqrt { 2 + 0.25 } + \sqrt {0.25} = 1.5 + 0.5 = 2$

$\sqrt{x+2} + \sqrt{x} = 2 \space \Rightarrow \text{Squaring both sides}$

$(x+2) + x + 2\sqrt{x(x+2)} = 4$

$2x + 2 + 2\sqrt{x^{2} + 2x} = 4 \space \Rightarrow \text{Dividing both sides with 2}$

$x + 1 + \sqrt{x^{2} + 2x} = 2$

$\sqrt{x^{2} + 2x} = 1 -x \space \Rightarrow \text{Squaring both sides again}$

$\cancel{x^{2}} + 2x = 1 - 2x + \cancel{x^{2}}$

$4x =1 \space \Rightarrow x = \dfrac{1}{4} = \boxed{0.25}$

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Note: Based on Pi Han Goh's comment below, we need to verify that $x= \dfrac{1}{4}$ is the only real number that can satisfy the equation. If $x= -\dfrac{1}{4}$ , we can clearly see that it's imaginary. And by checking, we have $x= \dfrac{1}{4}$ satisfies the equation.

Multiply the left hand side by (sqrt(x+2)-sqrt(x))/(sqrt(x+2)-sqrt(x)). We can get 2/(sqrt(x+2)-sqrt(x))=2, which is equivalently sqrt(x+2)-sqrt(x)=1. Solving it with the original equation, we get sqrt(x)=1/2, which means x=1/4.

$\begin{aligned} \sqrt{x+2} +\sqrt{x} & =2 \\(\sqrt{x+2})^2 & = (2-\sqrt{x})^2\\x+2 & = (2)^2 - 2(2)(\sqrt{x}) + (\sqrt{x})^2 \\x+2 & = 4 - 4\sqrt{x} + x \\x+4\sqrt{x} & = 2 +x \\4\sqrt{x} & = 2 \\\sqrt{x} & = \frac 24 \\x & = \frac 14 \\ \end{aligned}$

Verifying solution: $\sqrt{\frac 14+2} + \sqrt{\frac 14} = \frac 42 = 2$