Questions about a trick for solving quadratic equations quickly I just watched a video that teaches a trick to solve some quadratic equations faster:

Suppose we have $3x^2-152x+100=0$ It takes a lot of time to solve it
by finding discriminant because we have to calculate
$152^2$ and so on. we divide $3x^2$ by $3$ and multiply $100$ by $3$
and we get: $x^2-152x+300=0$ we can solve it easily by factoring
$(x-150)(x-2)=0$ then we divide the roots by $3$ so the roots of
original quadratic are $\frac{150}3$ and $\frac23$

It is the first time I see this trick. so is it a known method?
And How we can prove this method works mathematically?
 A: Nice trick, I never saw it.
$$ax^2+bx+c=0\iff a^2x^2+abx+ac=0\iff (ax)^2+b(ax)+ac=0.$$
So you solve for $ax$ and divide by $a$.

You can combine with another trick that I like, when $b$ is even or a factor $2$ can be pulled out (this is a frequent situation):
$$x^2+2bx+c=0\iff x=-b\pm\sqrt{b^2-c}.$$
A: Sure, your method can be proved to be correct.
Consider the quadratic - $ax^{2} +bx+c=0$ where $a$ is assumed to be non-zero. It's roots are given by the quadratic formula-
\begin{equation*}
\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}
\end{equation*}
And by your method, the quadratic changes to $x^{2}+bx+ac=0$.
It's roots are -
\begin{equation*}
\frac{-b\pm \sqrt{b^{2} -4ac}}{2}
\end{equation*}
You propose to divide the roots by $a$, so that just means they are the roots of the original equation.
Nice trick, but is it really feasible?
A: I have known a cousin of this trick for long, not sure of where from though.  Essentially like in factoring method, you multiply $3\times100$ and split the product $300$ into two so that the sum is $-152$, like so
$$3x^2-150x - 2x +100 = 3x(x-50)-2(x-50)=(3x-2)(x-50)$$

In general the logic here is that $$(px+q)(rx+s) = prx^2+(ps+qr)x+qs$$
so if there is a nice factoring of the quadratic, the middle term coefficient should be a sum of two whose product is $pqrs$.  One then considers the possible ways to split it back.
A: This trick is correct. But, unfortunately I cannot call it "nice".
I will try to show you the reason for this.

First alternatively, you can easily prove this  trick using Vieta's formula.
$$ax^2+bx+c=0$$
$$x_1+x_2=-\frac ba, ~ x_1x_2=\frac ca$$
Then,
$$y^2+by+ac=0$$
$$y_1+y_2=-b, ~ y_1y_2=ac$$
Putting $$x_1=\frac {y_1}{a},~ x_2=\frac{y_2}{a}$$
You get,
$$x_1+x_2=-\frac ba, ~ x_1x_2=\frac ca$$
which is correct.

I did not find this trick useful specifically for this equation for the following reason:
Observation
The implementation of this trick is based on the following observation:
$$150+2=152, ~ 150\times 2=300.$$
Isn't it?
But, this trick already works in the original equation:
$$\begin{align}3x^2-152x+100=0\end{align}$$
$$\iff 3x^2-150x-2x+100=0.$$
It is likewise easy to observe:
$$\frac{150}{3}=50, ~ \frac{100}{2}=50$$
This fact is equivalent to observe that
$$150+2=152, ~ 150\times 2=300.$$
Finally we have,
$$\begin{align}3x^2-152x+100=0
&\iff 3x^2-150x-2x+100 \\
&\iff 3x(x-50)-2(x-50) \\
&\iff (3x-2)(x-50).\end{align}$$
Which one is easy?  Make up your own mind.

Also, don't forget about the "half-discriminant" method.
If $$ax^2+2kx+c=0, a≠0$$ then
$$x_{1,2}=\dfrac{-k±\sqrt{k^2-ac}}{a}.$$
A: Let the original equation be $ax^2+bx+c=0$.
Let us divide degree $2$ term by $a$ and multiply constant term by $a$.
Then we get $x^2+bx+ac=0$.
The roots of this are $$x={{-b±\sqrt{b^2-4ac}}\over 2}$$
Dividing the roots by $a$, we get $${x\over a}={{-b±\sqrt{b^2-4ac}}\over 2a}$$
which are the roots of the original equation. Hence proved.
