In short

If \alpha and \beta are the roots of ax^2 + bx + c = 0, then \alpha + \beta = -b/a and \alpha\beta = c/a. These are Vieta's formulas. They let you compute the sum and product of the roots — and from those, any symmetric expression in \alpha and \beta — directly from the coefficients, without ever finding the roots themselves.

Solve x^2 - 5x + 6 = 0. The roots are 2 and 3. Now notice two things about them:

2 + 3 = 5 \qquad 2 \times 3 = 6

The sum of the roots, 5, is exactly the coefficient of x with its sign flipped. The product of the roots, 6, is exactly the constant term. That is not a coincidence — it works because a = 1 here, so -b/a = 5 and c/a = 6.

Try another one: 2x^2 + 3x - 5 = 0, whose roots are 1 and -5/2. Check:

1 + \left(-\tfrac{5}{2}\right) = -\tfrac{3}{2} = -\frac{b}{a} \qquad 1 \times \left(-\tfrac{5}{2}\right) = -\tfrac{5}{2} = \frac{c}{a}

Again it works. The sum and product of the roots are encoded in the coefficients of the equation — hiding in plain sight, available without any solving at all.

This relationship holds for every quadratic equation, not just ones with nice roots. Even when the roots are irrational like 2 \pm \sqrt{3}, their sum and product come out as clean rational numbers that you can read directly from a, b, c. That is the surprise, and it is the subject of this article.

Why it works

The proof is five lines. Start from the fact that any quadratic with roots \alpha and \beta can be written as

a(x - \alpha)(x - \beta) = 0

Expand the left side:

a\bigl[x^2 - (\alpha + \beta)x + \alpha\beta\bigr] = 0
ax^2 - a(\alpha + \beta)x + a\cdot\alpha\beta = 0

Now compare this with the original form ax^2 + bx + c = 0. The coefficients of x must match, and the constant terms must match:

b = -a(\alpha + \beta) \qquad c = a \cdot \alpha\beta

Solving for the sum and product:

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

That is the entire derivation. No quadratic formula needed, no completing the square — just expanding a product and matching coefficients.

Coefficient matching between factored form and standard formTwo rows showing the expanded factored form a times x squared minus alpha plus beta times x plus alpha beta on top, and the standard form a x squared plus b x plus c on the bottom, with arrows connecting matching coefficients to show that b equals negative a times alpha plus beta and c equals a times alpha beta. a·x² − a(α+β)·x + a·αβ = 0 match coefficients ↕ a·x² + b·x + c = 0 b = −a(α+β) c = a·αβ
The two forms of the same quadratic, with matching coefficients highlighted. The coefficient of $x$ encodes the sum of the roots (with a sign flip), and the constant term encodes the product. All the information about the roots is already in the coefficients.

The formal statement

Vieta's formulas for a quadratic

If \alpha and \beta are the two roots of ax^2 + bx + c = 0 (with a \neq 0), then:

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

Equivalently, in the monic form x^2 + px + q = 0 (where a = 1): the sum of the roots is -p and the product is q.

A note on when the roots are complex: Vieta's formulas work even then. If the discriminant is negative and the roots are \alpha = u + vi and \beta = u - vi, then \alpha + \beta = 2u = -b/a (a real number) and \alpha\beta = u^2 + v^2 = c/a (also real). The sum and product are always real when a, b, c are real, regardless of whether the roots themselves are.

Building equations from roots

Vieta's formulas work in reverse. If someone tells you the roots are \alpha and \beta, you can write the equation without expanding anything:

x^2 - (\alpha + \beta)x + \alpha\beta = 0

This is the monic quadratic (with a = 1) whose roots are \alpha and \beta. Multiply through by any non-zero constant a if you want a different leading coefficient.

For instance, to find a quadratic whose roots are 3 and -7:

The equation is x^2 - (-4)x + (-21) = 0, which simplifies to x^2 + 4x - 21 = 0.

Quick check: (x - 3)(x + 7) = x^2 + 7x - 3x - 21 = x^2 + 4x - 21. Correct.

This technique is especially useful when the roots are given in terms of other roots. A common JEE problem: "If \alpha and \beta are roots of x^2 + 5x + 3 = 0, find the equation whose roots are 1/\alpha and 1/\beta." You will see this solved in Example 2.

Symmetric functions of roots

The most powerful application of Vieta's formulas is computing symmetric functions of the roots — expressions that stay the same if you swap \alpha and \beta — without knowing the roots individually.

The sum \alpha + \beta and the product \alpha\beta are the two basic symmetric functions. Every other symmetric expression can be built from them. Here are the most commonly needed ones:

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta
\alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = (\alpha + \beta)\sqrt{(\alpha+\beta)^2 - 4\alpha\beta}
\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}
\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}

Each of these reduces to an expression involving only \alpha + \beta and \alpha\beta — which you already know from -b/a and c/a. No solving required.

Dependency tree of symmetric functions built from sum and productA tree diagram. At the bottom are two boxes labelled alpha plus beta and alpha times beta, representing Vieta's formulas. Arrows point upward to alpha squared plus beta squared, then further to alpha cubed plus beta cubed. A separate arrow from both base boxes points to one over alpha plus one over beta. This shows how all symmetric expressions reduce to the sum and product. α + β = −b/a αβ = c/a α² + β² = (α+β)² − 2αβ 1/α + 1/β = (α+β)/αβ α³ + β³ = (α+β)³ − 3αβ(α+β)
Every symmetric function of the roots is built from the sum and the product — the two quantities Vieta gives you for free. The arrows show dependencies: $\alpha^2 + \beta^2$ needs both the sum and the product, and $\alpha^3 + \beta^3$ needs all three (the sum, the product, and their combination).

An interactive Vieta's checker

Drag the point below to change b in x^2 + bx + 6 = 0. The readouts show -b/a (the sum of roots from Vieta's formula) and c/a = 6 (the product). Watch how the sum changes while the product stays fixed at 6 — the two roots slide along the parabola together, always multiplying to 6.

Interactive Vieta's formula explorer for x squared plus b x plus 6A horizontal slider from b equals negative 8 to b equals 8. A draggable red point controls b. Readouts display b, the sum of roots which equals negative b, the product of roots which is always 6, and the discriminant. As b changes, the sum changes but the product stays constant. −8 −4 0 4 8 drag to change b
Drag the point to vary $b$. The product stays fixed at $6$ (it depends only on $c/a$), but the sum $-b$ changes. When $|b| < 2\sqrt{6} \approx 4.90$, the discriminant is negative and the roots become complex — yet the sum and product remain real.

Two worked examples

Example 1: Find α² + β² for the roots of 2x² − 7x + 3 = 0

Step 1. Read off a, b, c.

a = 2, \quad b = -7, \quad c = 3

Why: these go straight into Vieta's formulas. No need to actually find \alpha and \beta.

Step 2. Apply Vieta's formulas.

\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}
\alpha\beta = \frac{c}{a} = \frac{3}{2}

Why: the sum and product are all the machinery you need. Every symmetric expression in \alpha and \beta can be written in terms of these two numbers.

Step 3. Use the identity \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta.

\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2 \cdot \frac{3}{2} = \frac{49}{4} - 3 = \frac{49 - 12}{4} = \frac{37}{4}

Why: this identity works because (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2. Subtracting 2\alpha\beta from both sides isolates \alpha^2 + \beta^2.

Step 4. Verify by solving. The roots are \alpha = 3 and \beta = 1/2 (factor as (2x - 1)(x - 3) = 0). Check: 9 + 1/4 = 37/4. Correct.

Result. \alpha^2 + \beta^2 = \dfrac{37}{4}.

Parabola y equals 2x squared minus 7x plus 3 with roots at one half and 3An upward-opening parabola crossing the horizontal axis at x equals one half and x equals 3. The vertex sits below the axis. Both roots are marked with red dots and labelled. x y 1 2 3 β = ½ α = 3 vertex α + β = 7/2 αβ = 3/2 α² + β² = 37/4
The parabola $y = 2x^2 - 7x + 3$ with its roots at $\beta = 1/2$ and $\alpha = 3$. Vieta's formulas give the sum $7/2$ and product $3/2$ by inspection. The sum of squares, $37/4$, follows from one algebraic identity — no need to solve the equation at all.

The key insight: you computed \alpha^2 + \beta^2 using only the sum and product of the roots. You could have solved the equation first, but Vieta's approach is faster, cleaner, and works even when the roots are ugly irrational numbers.

Example 2: If α, β are roots of x² + 5x + 3 = 0, find the equation whose roots are 1/α and 1/β

Step 1. Get the sum and product of the original roots from Vieta's formulas.

\alpha + \beta = -5, \qquad \alpha\beta = 3

Why: for a monic quadratic x^2 + px + q = 0, the sum is -p and the product is q. Here p = 5, q = 3.

Step 2. Compute the sum and product of the new roots 1/\alpha and 1/\beta.

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-5}{3}
\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{3}

Why: combining fractions over a common denominator gives the sum of reciprocals in terms of the original sum and product. This is the technique — express the new symmetric functions in terms of the old ones.

Step 3. Build the new equation using the formula x^2 - (\text{sum})x + (\text{product}) = 0.

x^2 - \left(-\frac{5}{3}\right)x + \frac{1}{3} = 0
x^2 + \frac{5}{3}x + \frac{1}{3} = 0

Step 4. Clear fractions by multiplying through by 3.

3x^2 + 5x + 1 = 0

Result. The equation whose roots are 1/\alpha and 1/\beta is 3x^2 + 5x + 1 = 0.

Two parabolas showing the original equation and the reciprocal-roots equationTwo parabolas plotted on the same axes. The wider one represents x squared plus 5x plus 3 with roots near negative 4.3 and negative 0.7. The narrower one represents 3x squared plus 5x plus 1 with roots near negative 1.4 and negative 0.2 — the reciprocals of the original roots. Both cross the negative x-axis. x y −1 −2 −3 −4 x² + 5x + 3 3x² + 5x + 1 α ≈ −4.30 β ≈ −0.70 1/α ≈ −0.23 1/β ≈ −1.43
The original equation (solid curve) has roots $\alpha$ and $\beta$ on the far left. The new equation (red curve) has roots $1/\alpha$ and $1/\beta$ — closer to zero, because reciprocals of large numbers are small. Vieta's formulas let you build the new equation without ever computing the roots explicitly.

Notice a pattern: the new equation 3x^2 + 5x + 1 = 0 has the same coefficients as the original x^2 + 5x + 3 = 0, but reversed. That is not a coincidence — if \alpha is a root of ax^2 + bx + c = 0, then 1/\alpha is a root of cx^2 + bx + a = 0, which you get by writing the original coefficients in reverse order. The Vieta's approach recovers this structural fact naturally.

Forming equations from given conditions

A very common exam question: "Form a quadratic equation whose roots satisfy some condition." The recipe is always the same.

  1. Compute the sum of the new roots (call it S).
  2. Compute the product of the new roots (call it P).
  3. The equation is x^2 - Sx + P = 0 (or multiply by a constant to clear fractions).

Here is a table of standard transformations. If \alpha and \beta are roots of ax^2 + bx + c = 0, and you want the equation whose roots are:

New roots Sum (S) Product (P)
\alpha^2, \beta^2 (\alpha+\beta)^2 - 2\alpha\beta = b^2/a^2 - 2c/a (\alpha\beta)^2 = c^2/a^2
1/\alpha, 1/\beta (\alpha+\beta)/(\alpha\beta) = -b/c 1/(\alpha\beta) = a/c
\alpha + k, \beta + k (\alpha+\beta) + 2k = -b/a + 2k \alpha\beta + k(\alpha+\beta) + k^2
k\alpha, k\beta k(\alpha+\beta) = -kb/a k^2\alpha\beta = k^2c/a

Each row comes from expressing the new sum and product in terms of the original \alpha + \beta and \alpha\beta, then substituting Vieta's values.

Applications beyond finding roots

Vieta's formulas show up in problems that look nothing like "solve a quadratic." Here are three patterns.

Checking answers. After you solve a quadratic, verify your answer by computing the sum and product of your roots and checking them against -b/a and c/a. If either check fails, you made an error. This is faster than substituting both roots back into the original equation.

Extracting symmetric information. If a problem asks for \alpha^2 + \beta^2, \alpha^3 + \beta^3, or \alpha/\beta + \beta/\alpha, you can compute the answer without solving the equation. Just express the desired quantity in terms of \alpha + \beta and \alpha\beta, then plug in -b/a and c/a.

Building equations from conditions. If you know the sum and product of the roots you want, you can write the equation directly. This is the reverse direction — from roots to equation — and Vieta's formulas are the bridge.

Common confusions

Going deeper

If you came here for the sum and product of roots, Vieta's formulas, and how to use them to build equations and compute symmetric functions, you have all of it — you can stop here. What follows connects these ideas to higher-degree polynomials and a structural perspective.

Vieta's formulas for cubics and beyond

For a cubic ax^3 + bx^2 + cx + d = 0 with roots \alpha, \beta, \gamma:

\alpha + \beta + \gamma = -\frac{b}{a}
\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}
\alpha\beta\gamma = -\frac{d}{a}

The pattern: the k-th symmetric function of the roots equals (up to sign) the ratio of the (k+1)-th coefficient to the leading coefficient. The signs alternate: -, +, -, +, .... For a degree-n polynomial, there are n such formulas, one for each elementary symmetric polynomial.

This is the general form of Vieta's insight: the coefficients of a polynomial are (up to signs and normalisation) the elementary symmetric functions of its roots. Every coefficient is a summary statistic of the roots.

Newton's identities

The symmetric functions \alpha^k + \beta^k (power sums) can be computed recursively from the elementary symmetric functions using Newton's identities. For a quadratic with sum s = \alpha + \beta and product p = \alpha\beta:

p_1 = s
p_2 = s \cdot p_1 - 2p
p_k = s \cdot p_{k-1} - p \cdot p_{k-2}

where p_k = \alpha^k + \beta^k. This gives you \alpha^k + \beta^k for any k without finding the roots. For instance, p_3 = s \cdot p_2 - p \cdot p_1 = s(s^2 - 2p) - ps = s^3 - 3ps, recovering the identity \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta).

The relationship to the discriminant

Vieta's formulas and the discriminant are closely connected. The discriminant is

D = b^2 - 4ac = a^2\bigl[(\alpha + \beta)^2 - 4\alpha\beta\bigr] = a^2(\alpha - \beta)^2

So \sqrt{D}/|a| = |\alpha - \beta| — the absolute difference of the roots. This is an asymmetric function (it changes sign if you swap \alpha and \beta), which is why the discriminant involves a square: D is symmetric, but \sqrt{D} is not. The discriminant is the bridge between the symmetric world (Vieta's formulas) and the asymmetric world (the individual roots).

Where this leads next