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Vieta's Formulas

What Are Vieta’s Formulas?

Section titled “What Are Vieta’s Formulas?”

Imagine if you could find relationships between the roots of a polynomial without actually calculating what those roots are. Sounds too good to be true? That’s exactly what Vieta’s formulas allow you to do.

Named after French mathematician François Viète, these elegant formulas create a direct connection between a polynomial’s roots and its coefficients. They’re incredibly powerful tools for solving complex problems involving roots—and often, you don’t even need to know what the roots actually are!

The Big Idea

  • Vieta’s formulas relate polynomial roots to coefficients
  • You can find sums and products of roots without solving the polynomial
  • These relationships work for quadratics, cubics, and higher-degree polynomials
  • This technique bypasses the need for lengthy root calculations

Quadratic Case: The Foundation

Section titled “Quadratic Case: The Foundation”

Let’s start with quadratic equations since they’re the most intuitive. Consider the quadratic with roots and .

Question
What are Vieta’s formulas for a quadratic ?
Answer

If and are the roots, then:

  • Sum:
  • Product:

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Building Intuition with an Example

Section titled “Building Intuition with an Example”

Let’s work through a concrete example to see how powerful this is.

Quadratic Roots - Using the Quadratic Formula

Section titled “Quadratic Roots - Using the Quadratic Formula”

Problem: Let and be the two roots of the quadratic . Using the quadratic formula, find: (a) (b) (c)

Solution:

For , we have , ,

Using the quadratic formula:

(a) The sum

(b) The product

(c) Using the identity:

Now let’s prove the general formula itself:

Vieta’s Formula for Quadratics - Proof

Section titled “Vieta’s Formula for Quadratics - Proof”

Problem: Let , , and be real numbers with , and let the quadratic have roots and . Prove that and .

Solution:

Since and are roots of , we can write:

Expanding the right side:

Comparing coefficients with :

  • Coefficient of : , so ✓
  • Constant term: , so ✓

Using Vieta’s Formulas Instead

Section titled “Using Vieta’s Formulas Instead”

Now here’s the magic: instead of using the quadratic formula, we can use Vieta’s formulas directly.

Quadratic Roots - Using Vieta’s Formulas

Section titled “Quadratic Roots - Using Vieta’s Formulas”

Problem: Let and be the two roots of . Use Vieta’s formulas to find: (a) (b) (c) (d)

Solution:

Using Vieta’s formulas: and

(a)

(b)

(c)

(d)

Notice how much faster this is than the quadratic formula!

More Challenging Quadratic Problems

Section titled “More Challenging Quadratic Problems”

Once you master the basics, Vieta’s formulas open up some beautiful algebraic manipulations:

Finding without finding and

Section titled “Finding without finding and ”

Problem: Let and be the two roots of . Find: (a) (b) (c) (d)

Solution:

From : and

(a)

(b)

We need

So:

(c)

(d)

Power Sums with Vieta’s Formulas

Section titled “Power Sums with Vieta’s Formulas”

Problem: Let and be roots of . Find: (a) (b) (c) (d) (e) (f)

Solution:

From : and

(a)

(b)

(c)

(d)

(e)

(f)

Cubic Polynomials: Extending the Pattern

Section titled “Cubic Polynomials: Extending the Pattern”

The beauty of Vieta’s formulas is that they extend to higher-degree polynomials. Let’s look at cubics:

Question
What are Vieta’s formulas for a cubic ?
Answer

If , , and are the roots, then:

  • Sum:
  • Sum of products of pairs:
  • Product:

Click to flip • Press Space or Enter

Proving Vieta’s Formulas for Cubics

Section titled “Proving Vieta’s Formulas for Cubics”

Vieta’s Formula for Cubics - Proof

Section titled “Vieta’s Formula for Cubics - Proof”

Problem: Let , , , and be real numbers with , and suppose the cubic has roots , , and . Prove that:

Solution:

Since , , and are roots:

Expanding

Then multiply by :

Comparing with :

  • Coefficient of : , so ✓
  • Coefficient of : , so ✓
  • Constant: , so ✓

Cubic Applications

Section titled “Cubic Applications”

Now let’s see how this works in practice with cubic polynomials:

Cubic Root Relationships

Section titled “Cubic Root Relationships”

Problem: Let , , and be the three roots of . Find: (a) (b) (c) (d) (e) (f)

Solution:

Using Vieta’s formulas for :

(a)

(b)

(c)

(d)

(e) Expanding:

(f)

Higher Power Sums in Cubics

Section titled “Higher Power Sums in Cubics”

Problem: Let , , and be the three roots of . Find: (a) (b) (c) (d) (e) (f)

Solution:

From : , ,

(a)

(b)

So:

(c)

We need

So:

(d) Converting to a common denominator:

Numerator

Using :

So the answer is

(e) - expand to:

(f) expands to:

Challenge: Cubic with Specific Properties

Section titled “Challenge: Cubic with Specific Properties”

Problem: Let , , and be the three roots of . Find: (a) (b) (c) (d) (e)

Solution:

From : , ,

(a)

(b) Since :

(c) requires

So:

(d) Using , , :

(e) Consider . We want .

This equals

Alternative: Expand using Vieta’s relations to verify:

Key Takeaways

Section titled “Key Takeaways”

Master These Concepts

  • Vieta’s formulas connect polynomial roots to coefficients algebraically
  • Quadratic case: For , we have and
  • Cubic case: For , three relationships connect sum, pairwise products, and full product to coefficients
  • Power of the method: You can find , reciprocal sums, and complex expressions without solving for and
  • The proof: Comes from factoring and comparing coefficients
  • Extension: Vieta’s formulas work for polynomials of any degree