Vieta's relations: roots and coefficients
Learning Objectives
- State and prove Vieta's relations \(x_1 + x_2 = -\tfrac{b}{a},\; x_1 x_2 = \tfrac{c}{a}\).
- Factor \(ax^2 + bx + c = a(x - x_1)(x - x_2)\) when \(\Delta \geq 0\), and explain the term double root.
- Form a quadratic equation with prescribed roots via \(x^2 - Sx + P = 0\).
- Find the roots of simple monic equations mentally, by sum and product.
- Determine the signs of the roots from \(S\) and \(P\) without solving.
- Check \(\Delta \geq 0\) before invoking any statement about the roots.
Solving backwards
Lesson 1 travelled in one direction: from the coefficients \(a, b, c\) to the roots. This lesson asks the opposite question: what do the roots know about the coefficients?
Look back at the garden equation \(x^2 - 10x + 21 = 0\), whose roots were \(3\) and \(7\). Now notice:
\[3 + 7 = 10 \qquad\text{and}\qquad 3 \cdot 7 = 21.\]
The sum and the product of the roots are sitting in plain sight inside the equation — the middle coefficient (with its sign flipped) and the constant term. Coincidence? Try another: \(x^2 - 5x + 6 = 0\) has roots \(2\) and \(3\), and indeed \(2 + 3 = 5\), \(2 \cdot 3 = 6\).
It is no coincidence, and the general statement — valid for any leading coefficient \(a\) — is one of the most useful facts about quadratic equations.
Vieta's relations
Let \(ax^2 + bx + c = 0\) (\(a \neq 0\)) be a quadratic equation with \(\Delta \geq 0\), and let \(x_1, x_2\) be its real roots (when \(\Delta = 0\), the double root is counted twice: \(x_1 = x_2\)). Then
\[S = x_1 + x_2 = -\frac{b}{a} \qquad\text{and}\qquad P = x_1 x_2 = \frac{c}{a}.\]
Both relations follow by direct computation from the quadratic formula of Lesson 1, \(x_{1,2} = \dfrac{-b \pm \sqrt{\Delta}}{2a}\), which is available precisely because \(\Delta \geq 0\).
For the sum, the square roots cancel:
\[x_1 + x_2 = \frac{-b + \sqrt{\Delta}}{2a} + \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}.\]
For the product, use the identity \((u + v)(u - v) = u^2 - v^2\) with \(u = -b\) and \(v = \sqrt{\Delta}\):
\[x_1 x_2 = \frac{(-b + \sqrt{\Delta})(-b - \sqrt{\Delta})}{(2a)^2} = \frac{b^2 - \Delta}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}.\]
When \(\Delta = 0\) both computations remain valid with \(x_1 = x_2 = -\tfrac{b}{2a}\).
Why "Vieta"?
The relations are named after François Viète (1540–1603), a French lawyer and royal advisor who did mathematics between duties — and transformed it. Viète was among the first to compute with letters standing for unknown and known quantities alike, which is exactly what makes a general statement such as \(x_1 + x_2 = -\tfrac{b}{a}\) expressible at all. Like many scholars of his time he published in Latin under a Latinized name, Franciscus Vieta — and it is this Latin form that survives in the phrase Vieta's relations (also called Vieta's formulas).
Dividing the equation by \(a\) puts the relations in their cleanest form: every quadratic equation can be written as
\[x^2 - Sx + P = 0,\]
where \(S\) is the sum and \(P\) the product of its roots. Mind the signs: minus the sum, plus the product.
Factoring the quadratic
Let \(ax^2 + bx + c = 0\) (\(a \neq 0\)) with \(\Delta \geq 0\) and roots \(x_1, x_2\). Then, for every real \(x\),
\[ax^2 + bx + c = a(x - x_1)(x - x_2).\]
In particular, when \(\Delta = 0\) the factorization reads \(a(x - x_1)^2\): the root appears twice as a factor. This is the promise of Lesson 1 kept — the reason \(x_1\) is called a double root.
Expand the right-hand side and substitute Vieta's relations:
\[a(x - x_1)(x - x_2) = a\bigl(x^2 - (x_1 + x_2)x + x_1 x_2\bigr) = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = ax^2 + bx + c.\]
Every step is an identity, valid for all \(x\).
Forming an equation from its roots
Conversely, let \(u\) and \(v\) be any two real numbers, and set \(S = u + v\), \(P = uv\). Then \(u\) and \(v\) are precisely the roots of the equation
\[x^2 - Sx + P = 0.\]
Consequently, two numbers are completely determined by their sum and product — up to their order.
For every real \(x\),
\[(x - u)(x - v) = x^2 - (u + v)x + uv = x^2 - Sx + P,\]
so the equation \(x^2 - Sx + P = 0\) says exactly \((x - u)(x - v) = 0\). A product is zero precisely when a factor is zero, so the solutions are exactly \(x = u\) and \(x = v\).
Vieta in both directions
Reading off sum and product. For \(2x^2 - 7x + 3 = 0\): without solving anything, \(S = -\tfrac{-7}{2} = \tfrac{7}{2}\) and \(P = \tfrac{3}{2}\). (Check by solving: \(\Delta = 49 - 24 = 25\), roots \(3\) and \(\tfrac{1}{2}\); indeed \(3 + \tfrac{1}{2} = \tfrac{7}{2}\) and \(3 \cdot \tfrac{1}{2} = \tfrac{3}{2}\).)
Forming an equation. Wanted: a quadratic equation with roots \(2\) and \(-5\). Here \(S = -3\) and \(P = -10\), so
\[x^2 - (-3)x + (-10) = 0, \qquad\text{that is,}\qquad x^2 + 3x - 10 = 0.\]
Solving in your head. Vieta turns simple monic equations into a puzzle: for \(x^2 - 5x + 6 = 0\), seek two numbers with sum \(5\) and product \(6\) — they are \(2\) and \(3\), and by the converse theorem they are exactly the roots. With practice this is faster than the formula whenever the roots are small integers.
The signs of the roots, without solving
Let \(ax^2 + bx + c = 0\) (\(a \neq 0\)) with \(\Delta \geq 0\), and let \(S = -\tfrac{b}{a}\), \(P = \tfrac{c}{a}\). Then:
If \(P < 0\), the roots have opposite signs — and in this case \(\Delta > 0\) holds automatically.
If \(P > 0\), the roots have the same sign: both positive when \(S > 0\), both negative when \(S < 0\).
If \(P = 0\), one root is \(0\) and the other equals \(S\).
The product of two real numbers is negative exactly when they have opposite signs, positive exactly when they have the same sign, and zero exactly when one of them is zero. Applied to \(x_1 x_2 = P\), this gives the three cases; it remains to fill in the details.
If \(P < 0\): then \(\tfrac{c}{a} < 0\), i.e. \(a\) and \(c\) have opposite signs, so \(ac < 0\) and \(\Delta = b^2 - 4ac \geq -4ac > 0\). The two distinct roots exist and their product is negative: opposite signs.
If \(P > 0\): the roots have the same sign, so their sum \(S\) carries that common sign — \(S > 0\) means both are positive, \(S < 0\) means both are negative. (Can \(S = 0\) happen here? No: dividing \(\Delta \geq 0\) by \(a^2\) gives \(S^2 - 4P \geq 0\), so \(S^2 \geq 4P > 0\).)
If \(P = 0\): then \(x_1 x_2 = 0\), so some root is \(0\), and the other equals \(x_1 + x_2 = S\).
Diagnosing signs at a glance
\(x^2 - 8x + 7 = 0\). Here \(P = 7 > 0\) and \(S = 8 > 0\), and \(\Delta = 64 - 28 = 36 \geq 0\): both roots are positive. (They are \(1\) and \(7\).)
\(3x^2 + 5x - 2 = 0\). Here \(P = -\tfrac{2}{3} < 0\): opposite signs, no need to check \(\Delta\) at all. (The roots are \(\tfrac{1}{3}\) and \(-2\).)
\(x^2 + 4x + 1 = 0\). Here \(P = 1 > 0\), \(S = -4 < 0\), \(\Delta = 16 - 4 = 12 \geq 0\): both roots are negative. (They are \(-2 + \sqrt{3}\) and \(-2 - \sqrt{3}\), both indeed negative since \(\sqrt{3} < 2\).)
Connection to Computer Science
Computers store numbers with limited precision, and the quadratic formula can lose that precision badly: when \(b^2\) is much larger than \(4ac\), the numerator \(-b + \sqrt{\Delta}\) subtracts two nearly equal quantities, and most of the reliable digits cancel away. Numerical software therefore computes only the "safe" root from the formula — the one where the two quantities are added, not subtracted — and recovers the other root from Vieta's product relation, \(x_2 = \dfrac{c}{a\,x_1}\), a single division that loses nothing. A four-century-old identity is standard practice inside modern numerical libraries.
Relations about roots that may not exist
Apply Vieta's relations, blindly, to \(x^2 + x + 1 = 0\): "the sum of the roots is \(-1\) and their product is \(1\)." The sentence is grammatical — but what is it about? Lesson 1 showed this equation has no real roots (\(\Delta = -3 < 0\)). A statement about the sum of things that do not exist asserts nothing.
This is why the hypothesis \(\Delta \geq 0\) appears in every theorem of this lesson: it is not decoration. The danger becomes real in problems with parameters — "find \(m\) so that the roots satisfy \(x_1 + x_2 = 3x_1x_2\)", say. Translating the condition through Vieta is only half the work; one must also impose \(\Delta \geq 0\), or the found \(m\) may describe an equation with no roots at all. Existence first, properties second — the same habit of rigor as in Lesson 1.
A final remark completes the picture: dividing \(\Delta \geq 0\) by \(a^2\) rewrites it as \(S^2 \geq 4P\). So the existence condition itself can be phrased in Vieta's own vocabulary: two real numbers with sum \(S\) and product \(P\) exist precisely when \(S^2 \geq 4P\).
Exercises
Exercise 1
Each equation below has real roots. Give \(S = x_1 + x_2\) and \(P = x_1 x_2\) without solving.
Exercise 2
Choose the monic quadratic equation whose roots are the given numbers.
Exercise 3
Solve each equation without the quadratic formula: find two numbers with the right sum and product.
Exercise 4
Decide whether each statement is true or false.
Exercise 5
Each equation has real roots. Using only \(S\) and \(P\), classify the signs of the roots.
Summary
Vieta's relations: when \(\Delta \geq 0\), the roots of \(ax^2 + bx + c = 0\) satisfy \(S = x_1 + x_2 = -\dfrac{b}{a}\) and \(P = x_1 x_2 = \dfrac{c}{a}\); a double root is counted twice.
Factorization: \(ax^2 + bx + c = a(x - x_1)(x - x_2)\); when \(\Delta = 0\) the factor \((x - x_1)\) appears twice — hence double root.
Forming equations: two numbers with sum \(S\) and product \(P\) are exactly the roots of \(x^2 - Sx + P = 0\) (mind the sign: minus the sum, plus the product). Such numbers exist precisely when \(S^2 \geq 4P\).
Signs without solving: \(P < 0\) — opposite signs (and \(\Delta > 0\) for free); \(P > 0\) — same sign, decided by \(S\); \(P = 0\) — one root is \(0\), the other is \(S\).
Existence first, properties second: never apply Vieta's relations before checking \(\Delta \geq 0\).