# Problem regarding the roots of a quadratic equation

I came across this problem on the internet:

If the roots of a quadratic equation $ax^2 + bx + b=0$ (where $a$ and $b$ are real numbers) are in the ratio $A:B$, then the value of $\sqrt{\dfrac{A}{B}} + \sqrt{\dfrac{B}{A}} +\sqrt{\dfrac{b}{a}} =$ __?

This was my approach to solving the problem:-

Let $At$ and $Bt$, $t \in \mathbb Z$ be the two roots of the quadratic equation. Then,

$(A+B)t = \dfrac{-b}{a}$; and

$(AB)t^2 = \dfrac{b}{a}$

Adding both the above equations, we get $t(ABt+A+B) = 0$. Now, if $t=0$, the two roots of the quadratic equation would be $0$, and the expression in the question is rendered meaningless owing to division by $0$. So, $(ABt+A+B) = 0 \implies t = \dfrac{-A-B}{AB}$.

Therefore, $\sqrt{\dfrac{b}{a}} = \sqrt{AB \dfrac{(-A-B)^2}{(AB)^2}} = \dfrac{-A-B}{\sqrt{AB}}$.

And so, $\sqrt{\dfrac{A}{B}} + \sqrt{\dfrac{B}{A}} +\sqrt{\dfrac{b}{a}} = \dfrac{A+B}{\sqrt{AB}}+ \dfrac{-A-B}{\sqrt{AB}} = 0$.

Is this correct? (I have a feeling that I've not correctly justified the argument that $t \neq 0$). Is there anywhere I can improve upon my writing? Please feel free to nitpick.

In your proof why should $t \in \mathbb{Z}$? In fact why should the roots be of the form $At$ and $Bt$? For example we know that $\frac{0.5}{0.25}=\frac{2}{1}$ but that doesn't mean $0.5$ is an integer multiple of $2$ and $0.25$ is the same integer multiple of $1$.
Let $\alpha$ and $\beta$ be the two roots. Then you are given that $$\frac{\alpha}{\beta}=\frac{A}{B}.$$ Using this we can write $$\sqrt{\dfrac{A}{B}} + \sqrt{\dfrac{B}{A}} +\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{\alpha}{\beta}} + \sqrt{\dfrac{\beta}{\alpha}} +\sqrt{\dfrac{b}{a}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{b}{a}}.$$