When reading about the quadratic equations, I came to the relation between the roots of the $p(x)$ with $a, b, c$.
Standard form of quadratic equation is: $ax^2 + bx + c$. Now, this is written in my textbook:-
In general, if $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = ax^2 + bx + c, a \ne 0$, then you know that $x – \alpha$ and $x – \beta$ are the factors of $p(x)$.
But how can $(x-\alpha)(x-\beta)$ be roots of $ax^2 + bx + c$ ?
$$(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta [= ax^2 - bx +c ] $$
and what is the problem in taking $(x + \alpha) (x + \beta)$ as roots. The only thing that changes is that zeroes will be negative i.e. $ x = -\alpha$ and $x = -\beta$. These negative zeroes actually make sense, because for an equation where all the values are positive, some negative value for x will be essential to make the $p(x) = 0$. Also, I don't see that there are much change in relation of zeroes and coefficients of x.
$$ (x+\alpha)(x+\beta) = kx^2 + k(\alpha + \beta)x + k\alpha\beta \\ \text{comparing } (ax^2 + bx+c) \text{ with }( kx^2 + k(\alpha + \beta)x + k\alpha\beta)\\ a=k, b=k(\alpha + \beta), c = k\alpha\beta\\ \text{So, }\alpha +\beta = \frac ba\\ \alpha\beta = \frac ca $$
Here the relations that $$\alpha+\beta = \frac{-b}{a}\\ \alpha\beta = \frac ca $$
are not much changed. There is a relation between the zeroes and the coefficients in either of the cases. Then why is it necessary to take roots as positive when it doesn't comply with the standard form of quadratic equation?