# Vieta's formulas for quadratic equation problem

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?

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.
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 $$x=\frac1a y$$ ...