Why can any quadratic equation be reduced to the product of its two factors involving the roots?

Question

I'm studying algebra and I'd like to know why can any quadratic equation be reduced to the product of its two factors.

And why $(x-α) (x-β)$ where $α$ and $β$ are its two roots?

And why is the sum of roots i.e. $α+β=-b/a$ and product of roots $\, αβ = c/a$?

Answer 1

Let an arbitrary quadratic expression be: $$q(x)=ax^2+bx+c$$ Define $$\alpha=\frac1{2a} (-b+\sqrt{b^2-4ac})$$ $$\beta=\frac1{2a} (-b-\sqrt{b^2-4ac})$$ Then a little algebra shows that $$\alpha+\beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}$$ $$a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x + a\alpha \beta = ax^2+bx+c$$

You should now be able to see the answer to all of your questions.

Answer 2

Wikipedia has a derivation of this factorization using completing the square on its quadratic formula page: https://en.wikipedia.org/wiki/Quadratic_formula#Method_3 It may also be enlightening to add and multiply the results of the quadratic formula:

$$\frac{-b + \sqrt{b^2-4ac}}{2a} + \frac{-b - \sqrt{b^2-4ac}}{2a} = \frac{-2b}{2a}$$

And:

$$\frac{-b + \sqrt{b^2-4ac}}{2a} \cdot \frac{-b - \sqrt{b^2-4ac}}{2a} = \frac{(-b)^2-(\sqrt{b^2-4ac})^2}{2a} = \frac{b^2-(b^2-4ac)}{4a^2}$$

Answer 3

Without using quadratic formula, we can do as follows.

Tool: If, $a≠0$ and $\alpha, \beta$ are the roots of $ax^2+bx+c$ then,

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

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We have,

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

$$\begin{align} ax^2+bx+c=&ax^2-a(\alpha+\beta)x+a\alpha\beta&\end{align}$$

$$b=-a(\alpha+\beta) ~ ,c=a\alpha\beta$$

$$\alpha+\beta=...~ \alpha\beta=...$$

I leave the rest of the last line to you.