Visual walkthrough — Vieta's formulas — sum and product of roots

Before we begin, three words we will use constantly — defined in plain English:

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Step 1 — A root means the graph touches zero

WHAT. Picture the quadratic as a curve: $y=a{x}^2+bx+c$. A root is exactly an $x$ where this curve crosses the horizontal line $y=0$.

WHY. We need to know what "$\alpha$ is a root" looks like before we manipulate it. It is not an abstract fact — it is a place on a picture where the curve meets the ground.

PICTURE. Below, the blue curve dips below the $x$-axis and comes back up. The two spots where it pierces the axis are marked $\alpha$ and $\beta$ — the roots.

Recall

What does it mean, on the graph, for $\alpha$ to be a root? ::: The curve $y=a{x}^2+bx+c$ passes through the point $(\alpha ,0)$ — it hits the $x$-axis there.

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Step 2 — Every root gives you a factor that is zero there

WHAT. Take the piece $(x-\alpha )$. This is a machine: feed it any $x$, it outputs $x$ minus the root. Notice that when $x=\alpha$, the output is $\alpha -\alpha =0$.

WHY. We want to rebuild the whole quadratic out of its roots. The key trick: a product is zero the moment any one of its parts is zero. So if we multiply $(x-\alpha )$ by $(x-\beta )$, the result is zero at both roots — exactly where our curve touches the axis.

PICTURE. Two straight lines. The line $y=x-\alpha$ crosses zero at $x=\alpha$; the line $y=x-\beta$ crosses zero at $x=\beta$. Each line "switches off" (hits $0$) at its own root.

$\underset{\text{zero when }x=\alpha }{\underbrace{(x-\alpha )}}\cdot \underset{\text{zero when }x=\beta }{\underbrace{(x-\beta )}}$

Each bracket is a light switch: whichever one turns off makes the whole product go dark (zero).

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Step 3 — Rebuild the quadratic from those factors

WHAT. A quadratic that is zero at $\alpha$ and $\beta$, and whose $x^2$ term has coefficient $a$, must be:

$y=a(x-\alpha )(x-\beta )$

WHY. Here is the reasoning behind the "must." Any quadratic is fixed by two things: where it crosses zero and how steep/stretched it is. The brackets nail the crossings; the number $a$ out front sets the vertical stretch. There is no other quadratic that agrees on both. This is why we are allowed to say $a{x}^2+bx+c$ equals $a(x-\alpha )(x-\beta )$ — same curve, two costumes.

PICTURE. The same blue parabola from Step 1, now drawn twice: once labelled $a{x}^2+bx+c$ (the "expanded costume") and once labelled $a(x-\alpha )(x-\beta )$ (the "factored costume"). They lie exactly on top of each other.

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Step 4 — Multiply the brackets out, watch each piece

WHAT. Expand $a(x-\alpha )(x-\beta )$ carefully. First multiply the two brackets using "each times each":

$(x-\alpha )(x-\beta )=\underset{x^2}{\underbrace{x\cdot x}}\underset{\text{from }x\cdot (-\beta )}{\underbrace{-\beta x}}\underset{\text{from }(-\alpha )\cdot x}{\underbrace{-\alpha x}}\underset{(-\alpha )(-\beta )}{\underbrace{+\alpha \beta }}$

WHY. We multiply because we want the equation back in the plain $x^2+\dots$ shape, so we can compare it to $a{x}^2+bx+c$. The two middle terms both contain a single $x$, so they collect together.

PICTURE. A $2\times 2$ grid (the "area box"). Rows labelled $x$ and $-\alpha$; columns labelled $x$ and $-\beta$. Each of the four cells is one product. The two shaded cells are the $x$-terms that merge.

Collect the two shaded middle cells:

$-\beta x-\alpha x=-(\alpha +\beta )x$

Read that term-by-term: the minus sign came from the two $(-\alpha )$ and $(-\beta )$; the thing that survived is exactly the sum $\alpha +\beta$. This is the birthplace of the negative sign. Now put the $a$ back:

$a(x-\alpha )(x-\beta )=a{x}^2\underset{\text{this will be }b}{\underbrace{-a(\alpha +\beta )}}x\underset{\text{this will be }c}{\underbrace{+a\alpha \beta }}$

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Step 5 — Match coefficients term by term

WHAT. Lay the two costumes side by side and equate matching powers of $x$:

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

• $x^2$ term: $a=a$ ✓ (nothing new — the stretch matches by design).
• $x^1$ term: $b=-a(\alpha +\beta )$.
• $x^0$ term (the constant): $c=a\alpha \beta$.

WHY. Because two identical curves cannot have different coefficients. Matching the $x$-line and the constant-line is the only place new information lives.

PICTURE. Three stacked colour bars — one bar per power of $x$ — with a bracket linking $b$ to $-a(\alpha +\beta )$ and $c$ to $a\alpha \beta$.

Now just divide by $a$ to isolate the root-quantities:

$$\boxed{\alpha +\beta =-\frac{b}{a}}\boxed{\alpha \beta =\frac{c}{a}}$$

Term-by-term reading: the sum is $b$ flipped in sign then scaled by $a$; the product is $c$ as-is scaled by $a$. That is the coefficient–root link in full — a special case of symmetric-polynomial structure. Compare with the Quadratic Formula: that gives the roots individually (needs a square root, discriminant and all); Vieta gives their sum and product with no square root at all.

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Step 6 — Edge case: what if $a\ne 1$? (non-monic)

WHAT. Everything above already carried the $a$, so no special magic — but you must divide by $a$ first in practice.

WHY. If you forget and read $\alpha +\beta =-b$ off a non-monic equation, you are secretly answering about the different equation $x^2+bx+c=0$, whose curve crosses the axis in different places.

PICTURE. Two parabolas with the same roots (same axis crossings) but different $a$: one flat, one steep. The $b$ and $c$ numbers differ, yet $-b/a$ and $c/a$ agree — showing why the $÷a$ is essential.

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Step 7 — Degenerate cases: repeated roots and complex roots

WHAT. Vieta never breaks, even when the picture stops crossing the axis twice.

• Repeated root ($\alpha =\beta$): the curve touches the axis at one point instead of crossing. Then $\alpha +\beta =2\alpha$ and $\alpha \beta ={\alpha }^2$. Still $-\frac{b}{a}$ and $\frac{c}{a}$.
• Complex roots (the curve never touches the axis): $\alpha$ and $\beta$ are a conjugate pair like $p+qi$ and $p-qi$. Their sum $2p$ and product $p^2+q^2$ are real numbers — so Vieta still returns real $-\frac{b}{a}$ and $\frac{c}{a}$. The imaginary parts cancel by design.

WHY. The derivation only used "the curve equals $a(x-\alpha )(x-\beta )$." That identity holds whether the roots are equal, distinct, or complex — the discriminant merely tells you which case you're in; it does not switch Vieta off.

PICTURE. Three parabolas in a row: two crossings (real distinct), one touch (repeated), floating above the axis (complex). Under each, its sum and product still equal $-b/a$ and $c/a$.

Recall

If a quadratic's graph never touches the $x$-axis, do Vieta's formulas still give real answers? ::: Yes. The roots are complex conjugates $p\pm qi$; their sum $2p$ and product $p^2+q^2$ are both real, so $-b/a$ and $c/a$ are real too.

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The one-picture summary

Everything above compressed: factored costume → area-box expansion → match coefficients → divide by $a$.

Recall Feynman retelling — the whole walkthrough in plain words

A quadratic curve dips down and comes back up, poking through the flat ground at two spots. Those spots are the roots. I build a machine $(x-\alpha )$ that switches off exactly at the first spot, and $(x-\beta )$ that switches off at the second. Multiply them and stretch by $a$, and I've rebuilt the very same curve — because a parabola is completely pinned down by where it crosses and how stretched it is. Now the same curve is written two ways: the plain $a{x}^2+bx+c$ and the multiplied-out brackets. Same curve means the numbers in front must match. When I multiply the brackets out in a little $2\times 2$ area box, the two middle cells both carry a minus sign, and together they spell $-(\alpha +\beta )$ — that's where the famous minus comes from. The corner cell $(-\alpha )(-\beta )$ turns into a plain plus, giving $+\alpha \beta$. Matching those to $b$ and $c$, then dividing by the stretch $a$, hands me the sum and the product without ever solving anything. And it doesn't care if the roots are the same, or complex — the "same curve" argument never used that.

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Links: Parent topic · Factoring Quadratics · Completing the Square · Quadratic Formula · Symmetric Polynomials · Polynomial Roots and Coefficients · Systems of Equations · Descartes' Rule of Signs