Need help with Alternative Factoring method

Question

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?

Answer 1

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 $\frac{5}{4}\pm\frac{7}{4}=\{\frac{-1}{2},3\}$.

Answer 2

You have to pay attention to the signs. In particular, if $r$ and $s$ are roots of the quadratic $$x^2 - bx + c,$$ then $r + s = b$ and $rs = c$. Notice that my signs are different for the coefficients of $b$ and $c$; whereas you have $x^2 + bx - c$, I have $x^2 - bx + c$. This is necessary in order to have $r + s = b$ and $rs = c$, because $$(x-r)(x-s) = x^2 - (r+s)x + rs.$$ Your particular example has $b = 5/2$, $c = -3/2$, for which $(r,s) \in \{(-1/2, 3), (3, -1/2)\},$ and we can take without loss of generality either solution as the roots are invariant with respect to permutation. You solved the case $b = 5/2$, $c = 3/2$.