Vieta's Formulas

r + s = -b, rs = c — work with roots without ever finding them.

The idea

Vieta's formulas read the sum and the product of a polynomial's roots directly off its coefficients, with no need to solve the equation.

Theorem (Vieta's formulas).

If the monic quadratic $x^{2} + bx + c$ has roots $r$ and $s$, then $r + s = -b \qquad \text{and} \qquad rs = c.$

In higher degrees the coefficients likewise give the sums of the roots taken one at a time, two at a time, and so on, with alternating signs.

The sum and the product determine every symmetric expression in the roots — every expression unchanged by swapping them — though never an individual root. For example, $r^{2} + s^{2} = (r+s)^{2} - 2rs$ and $\tfrac{1}{r} + \tfrac{1}{s} = \tfrac{r+s}{rs}$ are both built from the sum and the product alone.

Reach for Vieta's formulas whenever a problem asks for a symmetric quantity in the roots, above all when the roots are irrational, so that solving first would carry square roots through every step.

Ways to work on it

• Walkthrough. Read the sum and product of a quadratic's roots off its coefficients, then evaluate a symmetric expression without solving.
• Proof. See why the coefficients are the sums and products of the roots — for the cubic, and for any degree.
• Practice. Evaluate a random symmetric expression in the roots from the sum and product.
• Hardest. Apply Vieta's formulas to a cubic, or build a quadratic whose roots are shifts of another's.

Not sure where to start? Take the ten-question placement test.