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Problem 1
$$\sqrt{3+x}+3\sqrt{x}=5$$
We raise both sides with power of 2 to get "rid of" the square root.
We use the quadrat rule.
$$\sqrt{3+x}+3\sqrt{x}=5$$
$$\Downarrow$$

$$(\sqrt{3+x}+\sqrt{x})^2=5^2$$
$$\Updownarrow$$
$$(x+3)+6\cdot\sqrt{x+3}\cdot\sqrt{x}+x=25$$
We rewrite the expression
$$(x+3)+6\sqrt{x+3}\cdot\sqrt{x}+9x=25$$
$$\Updownarrow$$
$$10x-22+6\cdot\sqrt{x+3}\cdot\sqrt{x}=0$$
We divide both sides with 2, rewrite and square both sides.
$$10x-22+6\cdot\sqrt{x+3}\cdot\sqrt{x}=0$$
$$\Updownarrow$$
$$5x-11+3\sqrt{x+3}\cdot\sqrt{x}=0$$
$$\Downarrow$$
$$(3\sqrt{x+3}\cdot\sqrt{x})^2=(11-5x)^2$$
$$\Updownarrow$$
$$9x^2+9x=121-110x+25x^2$$
$$\Updownarrow$$
$$16x^2-137x-121=0$$
We guess a root, if there is a root $x\in\textbf{Z}$
It is natural to try with $1$ first and we see that $x=1$ is a root to the equation.
So in it follows that $$16x^2-137x-121=0$$ is divisible with $$(x-1)$$ and can be written as,
$$16x^2-137x-12=(x-1)p(x)+r(x)$$
Since $$(x-1)\mid 16x^2-137x-121$$ the rest polynomial $r(x)$ is $0$
By polynomial division we see that $$\frac{16x^2-137x-121}{x-1}=16x-121$$
Therefore $$16x^2-137x-121$$ can be factorised as $$(16x-121)(x-1)$$
$(16x-121)(x-1)=0$ has the roots $x_0=1$ and $x_1=\displaystyle\frac{121}{16}$
By testing the roots in the original equation $\sqrt{3+x}+3\sqrt{x}=5$ we see that only $x_0=1$ is a legitimate root. And we knew from the beginning that  $\sqrt{3+x}+3\sqrt{x}=5$ has only one root.
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