Foundations — Vieta's formulas — sum and product of roots

Before you can read that message, you need to own every single symbol the parent note throws at you. This page builds each one from absolute zero, in an order where nothing appears before it is explained.

0. The three arithmetic operations we assume

The parent note leans on ordinary arithmetic. We name these first so nothing later sneaks in unexplained.

With $+$, $-$, $\times$, and the power counter defined, every symbol below is now legal to use.

1. The equals-zero equation: $a{x}^2+bx+c=0$

Now the most basic character on stage.

What each letter is, in plain words:

• $x$ — the mystery number. Picture a slot waiting to be filled.
• $a$ — how "steep" the squared part is (the leading coefficient). Must not be $0$, or the $x^2$ vanishes and it stops being quadratic.
• $b$ — how much of the plain $x$ we add.
• $c$ — a fixed number added at the end (the constant term).
• $=0$ — we are asking: for which $x$ does the left side hit exactly zero?

The spots where the parabola crosses the ground line are the numbers we care about. They have a name.

2. Roots: $\alpha$ and $\beta$

The parent note names the two roots with Greek letters:

• $\alpha$ (alpha) — the first root, just a name for a number.
• $\beta$ (beta) — the second root.

A quadratic has at most two roots because its parabola can cross a horizontal line at most twice. But "at most two" hides three genuinely different situations — we must cover all of them.

3. The factor form: $(x-\alpha )(x-\beta )$

This is the heart of the whole topic, so we go slowly.

Why $(x-\alpha )$ and not $(x+\alpha )$? Because we want the bracket to become $0$ exactly when $x$ equals the root. Substitute $x=\alpha$ into $(x-\alpha )$: you get $\alpha -\alpha =0$ (subtraction of a number from itself gives zero).

4. Multiplying brackets out (expansion) and the FOIL move

To decode coefficients we must multiply the two brackets back together (using the multiplication defined in section 0). Multiplying out brackets is called expansion. The rule: every term in the first bracket must be multiplied by every term in the second.

$(x-\alpha )(x-\beta )=\underset{\text{First}}{\underbrace{x\cdot x}}\underset{\text{Outer}}{\underbrace{-\beta x}}\underset{\text{Inner}}{\underbrace{-\alpha x}}\underset{\text{Last}}{\underbrace{+\alpha \beta }}$

(Here $x\cdot x$ is another way to write the multiplication $x\times x=x^2$, and $\alpha \beta$ is the side-by-side shorthand for $\alpha \times \beta$.)

Now collect the two middle terms, because both contain a single $x$: $=x^2-(\alpha +\beta )x+\alpha \beta$

Look at what popped out on its own:

• The coefficient of $x$ became $-(\alpha +\beta )$ — the sum of the roots, with a minus.
• The constant became $\alpha \beta$ — the product of the roots.

That is Vieta's formula appearing before your eyes, purely from multiplying brackets.

5. Sum ($\alpha +\beta$) and product ($\alpha \beta$)

Two more plain-English symbols, both built from the operations of section 0:

• $\alpha +\beta$ — add the two roots.
• $\alpha \beta$ (two letters touching) — multiply the two roots, i.e. $\alpha \times \beta$.

Matching our expansion $x^2-(\alpha +\beta )x+\alpha \beta$ against the general form gives both Vieta relations. Let us state the second one plainly, since it is easy to miss:

6. The fraction bar $\frac{b}{a}$ and dividing by the leading coefficient

The parent writes $\alpha +\beta =-\frac{b}{a}$. The fraction bar just means divide: $\frac{b}{a}=b÷a$.

7. Powers again and the square identity: ${\alpha }^2+{\beta }^2$

We already met the power counter in section 0 (${\alpha }^2=\alpha \times \alpha$). The examples in the parent use it inside an identity: ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

8. How it all connects

The diagram below is a prerequisite map: read it top to bottom. Start at the equation (top). It defines its roots (node B). The roots let you write the factor form (C). Expanding that form (D) and collecting the middle terms (E) drops two facts into your lap: the sum equals $-b/a$ (F) and the product equals $c/a$ (G) — but only after you divide by $a$ to make things monic (H feeds both F and G). Finally, the square identity (I) combined with the sum and product produces ${\alpha }^2+{\beta }^2$ (J), and F together with G form the complete Vieta decoder (K).

If the diagram fails to render in your reader, the paragraph above it says exactly the same thing in words — no information is lost.

Related vault notes

The machinery here leans on and feeds into: Quadratic Formula, Factoring Quadratics, Completing the Square, Discriminant, Polynomial Roots and Coefficients, Symmetric Polynomials, Systems of Equations, and Descartes' Rule of Signs. Return to the parent: Vieta's formulas.

Equipment checklist

Test yourself — cover the right side and answer out loud.