For example, when finding the roots of this quadratic equation:
$x^{2}+8x-9=0$
I would write
$x^2-x+9x-9=0$
Then write that expression as the product of two linear expressions:
$(x+9)(x-1)=0$
Then I solve the two linear equations for the roots of the quadratic equation which are $x_1=-9$ and $x_2=1$
In that method I had to find two numbers such that their product is -9 and their sum is 8. This is straightforward enough but I am curious.
In general, this is the task:
Let $px^2+qx+r=0$
Find numbers $\alpha,\beta$ such that $\alpha\beta=pr$ and $\alpha+\beta=q$
Does a general method exist for finding such pairs $\alpha,\beta$? I know that I can list factors and then inspect them to find a pair of solutions to those equations, but I am interested in whether I can do this a different way.
I have read the similar questions I could find on this website, and the answers only contained the trial-and-error method.
In case of quadratic equations, the kind that high school student come across most frequently, hit-n-trial is often the best approach. And if you are just interested in the roots, then of course, you can always use the quadratic formula or a calculator.
However, it appears you are more focused on a different way of finding factors of a number that add to another number. So here's one you could use.
We have
$$\begin{align*} \alpha + \beta &= s \hspace{1.5cm} - (1) \\[0.3cm] \alpha \beta &= p \\[0.3cm] \end{align*}$$
Now,
$$ (\alpha + \beta)^2 - 4\alpha\beta = (\alpha - \beta)^2 $$
So,
$$ \alpha - \beta = \sqrt{s^2 - 4p} \hspace{1.5cm} - (2) $$
I have ignored the negative root.$^{*}$
That gives us two equations in $\alpha$ and $\beta$ that can be solved easily to get the two numbers.
$^{*}$ Taking the negative root will give you the same numbers, just in the opposite order.