Vieta's Formulas for Quadratics — Derivation and Basic Uses

What Are Vieta’s Formulas?

Vieta’s formulas are one of the most elegant shortcuts in algebra. They reveal a hidden connection between the roots of a polynomial and its coefficients—and the best part? You don’t even need to find the actual roots to use them.

Imagine being asked to find where and are roots of some quadratic. Most people would calculate the roots first using the quadratic formula, then square and add them. But Vieta’s formulas let you skip the messy root-finding entirely and jump straight to the answer.

The Core Idea

Vieta’s formulas relate roots to coefficients without calculating the roots themselves
For quadratics: sum and product of roots depend only on the coefficients
This unlocks elegant solutions to otherwise complicated problems
The same principle extends to cubics, quartics, and higher-degree polynomials

Deriving Vieta’s Formulas for Quadratics

Let’s build this from the ground up. Suppose we have a quadratic equation:

where , and let’s call the two roots and .

Starting with the Quadratic Formula

We know the roots are:

Now let’s find what and equal.

Sum of the Roots

Adding the two root expressions:

Notice how the discriminant term cancels out beautifully. The sum depends only on and !

Product of the Roots

Now for the product, we use the difference of squares pattern :

Again, something wonderful happens—we get a simple expression depending only on and .

Vieta’s Formulas for Quadratics

For the quadratic with roots and :

That’s it. That’s the whole power of Vieta’s formulas for quadratics.

Applying Vieta’s Formulas: Quadratic Examples

Let’s see how powerful these formulas are when solving problems.

Warm-Up: Direct Application

Problem: Let and be the two roots of . Find:

(a)

(b)

(c)

Solution:

From , we have , , .

(a)

(b)

(c)

Building Expressions from Roots

Problem: Let and be the two roots of . Find:

(a)

(b)

(c)

(d)

Solution:

From : and

(a)

(b) First find

(c)

(d)

Building Higher Powers of Roots

One of the trickiest applications of Vieta’s formulas is finding sums of higher powers like or . But there’s a systematic approach using recurrence relations.

Powers and Recurrence

Problem: Let and be the two roots of . Find:

(a)

(b)

(c)

Solution:

From : and

(a)

(b)

(c) Using the recurrence :

, ,

✓ (matches part b)

Extending to Cubic Polynomials

The beauty of Vieta’s formulas is that they generalize. For cubic polynomials, we get three relationships instead of two.

Vieta’s Formulas for Cubics

For the cubic with roots , , and :

The pattern is elegant: coefficients determine the elementary symmetric polynomials in the roots.

Cubic Application

Problem: Let , , and be the three roots of . Find:

(a)

(b)

(c)

(d)

(e)

(f)

Solution:

From :

(a)

(b)

(c)

(d)

(e)

(f)

Advanced Cubic Problems

Once you master the basics, cubic problems become fascinating puzzles that reward creative thinking.

Power Sums in Cubics

Problem: Let , , and be the three roots of . Find:

(a)

(b)

(c)

Solution:

From : , ,

(a)

(b) Since each root satisfies the cubic:

(c)

Numerator:

Denominator:

Complex Cubic Expressions

Problem: Let , , and be the three roots of . Find:

(a)

(b)

(c)

(d)

(e)

Solution:

From : , ,

(a)

(b) Since , we have

(c) Following previous pattern: numerator , denominator

(d)

(e) Consider that satisfies . We want .

This is the polynomial evaluated at points where .

and

The product

After careful calculation: The answer is

Key Insights and Patterns

Let me highlight the powerful patterns that emerge as you work with Vieta’s formulas:

Master Patterns

Sum of roots: Always
Products and sums: Can be combined to build more complex expressions
Recurrence relations: For power sums , use
Polynomial substitution: Since roots satisfy the original equation, you can use that to your advantage
Special cases matter: When , calculations simplify dramatically

Question

For

with roots

and

, what are Vieta’s formulas?

Answer

and

Click to flip • Press Space or Enter

Question

How do you find

using Vieta’s formulas?

Answer

Click to flip • Press Space or Enter

Question

What’s the identity for

in terms of elementary symmetric polynomials?

Answer

Click to flip • Press Space or Enter

Why This Matters

Vieta’s formulas aren’t just clever tricks for competition math. They appear throughout:

Algebra: Understanding polynomial structure and behavior
Precalculus: Building intuition for how equations relate to their solutions
Calculus: Analyzing critical points and understanding function behavior
Engineering and Physics: Many physical systems are modeled by polynomial equations
Computer Science: Polynomial evaluation and root-finding algorithms

Most importantly, Vieta’s formulas teach a fundamental principle: sometimes the relationships are more powerful than the individual elements. You can solve problems about roots without ever computing them explicitly.

Practice and Exploration

The exercises in this lesson progress from straightforward applications to increasingly sophisticated problems. As you work through them:

First practice direct application—just plug values into the formulas
Then tackle algebraic manipulations—combining formulas creatively
Finally, attack complex expressions—where you must think strategically about which relationships to use

The more problems you solve, the more you’ll develop intuition for which approach works best.

Learning Checklist

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Remember: mastering Vieta’s formulas opens up a whole new approach to polynomial problems. The more you use them, the more elegant and efficient your solutions become.