The above question is in reference to the process of solving the following problem.
Question: If the arithmetic means of two positive numbers $a$ and $b$, where $(a>b)$ is twice their G.M then prove that $a:b=(2+\sqrt{3}):(2-\sqrt{3})$.
Solution: (As $a>b$),
Let $(A=A.M=\frac{a+b}{2}$ and $G=G.M=\sqrt{ab})$ Therefore, the equation whose roots are $a$ and $b$ is
$x^2−(a+b)x+ab=0$
$x^2−2Ax+G^2=0$
$x^2−2(2G)x+G^2=0 $
$x^2−4Gx+G^2=0 $ Therefore, $x=(2\pm \sqrt{3})G$. Hence the result.
The first equation ( or the first step) in above solution process is true only if $a$ and $b$ are roots of a quadratic equation. How do we know from the question that they are roots?
Given any field $\Bbb{F}$ (e.g. $\Bbb{R}$ in this case), and any two $a, b \in \Bbb{F}$, we can always find a quadratic equation (with coefficients in $\Bbb{F}$) whose solutions are precisely $a$ and $b$. In particular, $$(x - a)(x - b) = 0.$$ This is simple to verify: note $x = a$ and $x = b$ are solutions, and the zero product property states that any solution $x$ must satisfy $x - a = 0$ or $x - b = 0$.
Expanding this equation yields the first step: $$x^2 - (a + b)x + ab = 0,$$ which once again, is equivalent to $x = a$ or $x = b$.
This can be done for any two $a, b$ in the field. Note that, if we have $\Bbb{F} = \Bbb{R}$, then there's no guarantee that the coefficients $-(a + b)$ and $ab$ are going to be integers or rational, but at the same time, there is no requirement that they should be. So, if you're worried about questions of transcendental numbers, don't be; it's ok if the coefficients are not rational.
The argument shows that, \begin{align*} x = a \text{ or } x = b &\iff x^2 - (a + b)x + ab = 0 \\ &\iff \ldots \\ &\iff x = (2 + \sqrt{3})G \text{ or } x = (2 - \sqrt{3})G. \end{align*} Since $a > b > 0$ and hence $G > 0$, we know that $a = (2 + \sqrt{3})G$ and $b = (2 - \sqrt{3})G$, hence the ratio follows.