Equations with two roots

With two equal roots it is enough to set the radicands equal and require one of them to be nonnegative, because the equality of the radicands makes the other condition automatic. The first-degree equation gives $x=4$, which satisfies $4+13=17\ge 0$. Checking by substitution confirms the result, since both radicands equal $17$ and both sides equal $\sqrt{17}$.

Of the two radicands, it is convenient to require the nonnegativity of $2x$, which is the simpler one. The equality of the radicands gives the second-degree equation $x^2-2x-3=0$, with roots $3$ and $-1$; the negative root does not satisfy the condition $x\ge 0$ and is discarded, and indeed for $x=-1$ the radicand $2x=-2$ is negative and the right-hand side is not defined. Checking by substitution confirms the remaining solution: for $x=3$ both radicands equal $6$.

The two roots have different radicands and a constant is added to the right-hand side, so the rule of equal radicands does not apply directly and both sides are squared. Squaring is legitimate because both sides are nonnegative where $x\ge 0$. In the square of the right-hand side the cross term $2\sqrt{x}$ appears; the $x$ terms cancel and what remains is an equation in the root $\sqrt{x}$ alone, which equals $2$, hence $x=4$ with a second squaring, again between nonnegative sides. Checking by substitution confirms the result: $\sqrt{9}=3$ and $1+\sqrt{4}=3$.