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I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it. Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divide roots by $a$.

For example: $$-6x^2+7x+5=0$$ I solve $$ y^2 + 7y -30 = 0\\ y_1=-10\\ y_2=3 $$ And then divide the roots: $$ x_1=\frac{-10}{-6}=\frac{5}{3}\\ x_2=\frac{3}{-6}=-\frac{1}{2} $$ Which gives me a correct answer. But I want to know why it's so. For now what I figured out is only that: $$ \text{for } a\neq 0 \text{ :}\\ ax^2 + bx + c = 0 \Leftrightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0\\ x_1 + x_2 = -\frac{b}{a} \Leftrightarrow a(x_1 + x_2) = -b\\ x_1 \cdot x_2 = \frac{c}{a} \Leftrightarrow a(x_1 \cdot x_2) = c\\ $$ $$ y^2 + by + ca = 0\\ y_1 + y_1 = -b\\ y_1 \cdot y_2 = ca \Leftrightarrow c = \frac{y_1 \cdot y_2}{a}\\ $$ $$ a(x_1 + x_2) = y_1 + y_1 \Leftrightarrow x_1 + x_2 = \frac{y_1}{a} + \frac{y_2}{a}\\ a(x_1 \cdot x_2) = \frac{y_1 \cdot y_2}{a} \Leftrightarrow x_1 \cdot x_2 = \frac{y_1}{a} \cdot \frac{y_2}{a}\\ $$ Any ideas?

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Take the original equation $ax^2+bx+c=0$ then multiply by $a$ to obtain $$a^2x^2+abx+ac=0$$Now set $y=ax$ so that $$y^2+by+ac=0$$ and you are done.

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  • $\begingroup$ Damn, that was so easy. Thank you so much! $\endgroup$ – Zekfad Apr 24 '20 at 21:49
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    $\begingroup$ @Zekfad That trick of changing the variable can often be used with Vieta to shorten calculations (cubics and quartics are problems often set in this kind of area) $\endgroup$ – Mark Bennet Apr 25 '20 at 9:47
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By the quadratic formula the roots of $aX^2+bX+c$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

The roots of $Y^2+bY+ca$ are $y=\frac{-b\pm \sqrt{b^2-4ca}}{2}$...

We see that $x=\frac1a y$ ...

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