<s>[INST] Need help with Alternative Factoring method I was working on some factoring, as I have always been terrible at it, when I found 3B1B's video on an easier method. There's a TL:DR at the bottom if you're familiar. The basics are as follows:
Imagine the graph of a quadratic. $x^2 - 1$ for example. It's got 2 roots $r$ and $s$ the same distance $d$ apart from a midpoint $m$.
The method only works for equations that look like $x^2 +b-c=0$, so if you've got an $a$, scale everything else down by dividing everything by $a$.
So now you've got that:
$r+s=b$ and $r \cdot s = c$.
We also know that:
$r=m-d$ and $s=m+d$.
We can do some algebra to the above to realize the following:
$m=\frac{-b}{2}$
and
$d = \sqrt{m^2-c}$
So neat! we've got a simple way to factorize! Except something must turn out wrong, because I've gotten a wrong result.
$2x^2 -5x -3 = 0$
scales down to
$x^2 -\frac{5}{2}x -\frac{3}{2} = 0$
And after some crunching, we get that $r$ and $s$ are $1$ and $\frac{6}{4}$, respectively.
That's not accurate. The right answers are $-\frac{1}{2}$ and $3$. what gives?
TL;DR:
Using alternative quadratic method on
$x^2 -\frac{5}{2}x -\frac{3}{2} = 0$ gets me the x-intercepts at $x=-3$. Why? and how can I make sure this doesn't happen again?
 [/INST] Given
$$x^2+bx+c=x^2-\frac{5}{2}x-\frac{3}{2}$$
We get $m=\frac{-b}{2}=\frac{5}{4}$ and $$d=\sqrt{m^2-c}=\sqrt{\frac{25}{16}+\frac{3}{2}}=\sqrt{\frac{49}{16}}=\frac{7}{4}$$
Therefore our roots are $\frac{5}{4}\pm\frac{7}{4}=\{\frac{-1}{2},3\}$.
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