Roots of polynomial equations

Welcome to the World of Polynomials!

In this chapter, we are going to explore the secret relationship between the roots of an equation (the answers you get when you solve it) and its coefficients (the numbers in front of the variables).

Think of a polynomial equation like a recipe. The coefficients are the ingredients listed on the package, and the roots are the finished cake. Even if we haven't "baked" the cake yet, we can tell a lot about its flavor just by looking at the ingredients! This is a vital skill in Further Mathematics because it allows us to solve complex problems without always having to find the exact numerical values of the roots first.

1. The Basics: Roots and Coefficients

Whether you are dealing with a quadratic (degree 2), a cubic (degree 3), or a quartic (degree 4) equation, there is a beautiful, repeating pattern in how the roots relate to the coefficients.

Quadratic Equations (Degree 2)

For the equation \(ax^2 + bx + c = 0\), let the roots be \(\alpha\) (alpha) and \(\beta\) (beta).

1. Sum of the roots: \(\alpha + \beta = -\frac{b}{a}\)
2. Product of the roots: \(\alpha\beta = \frac{c}{a}\)

Cubic Equations (Degree 3)

For the equation \(ax^3 + bx^2 + cx + d = 0\), let the roots be \(\alpha\), \(\beta\), and \(\gamma\) (gamma).

1. Sum of roots: \(\alpha + \beta + \gamma = -\frac{b}{a}\)
2. Sum of roots taken in pairs: \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\)
3. Product of roots: \(\alpha\beta\gamma = -\frac{d}{a}\)

Quartic Equations (Degree 4)

For the equation \(ax^4 + bx^3 + cx^2 + dx + e = 0\), let the roots be \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\) (delta).

1. Sum of roots: \(\sum \alpha = -\frac{b}{a}\)
2. Sum of products in pairs: \(\sum \alpha\beta = \frac{c}{a}\)
3. Sum of products in triples: \(\sum \alpha\beta\gamma = -\frac{d}{a}\)
4. Product of all roots: \(\alpha\beta\gamma\delta = \frac{e}{a}\)

Quick Review Tip: Notice the pattern of the signs! It always starts with a negative for the sum, then alternates: \(-, +, -, +\). The denominator is always the leading coefficient \(a\).

Did you know? These relationships are known as Vieta's Formulas, named after the French mathematician François Viète. He was one of the first people to use letters to represent numbers in math!

Key Takeaway: The sum of roots is always \(-\frac{second\ coefficient}{first\ coefficient}\), and the product of roots involves the last coefficient. The signs always alternate.

2. Symmetric Functions of Roots

A symmetric function is an expression where, if you swap any two roots (like swapping \(\alpha\) and \(\beta\)), the expression stays exactly the same. In your exam, you will often be asked to find the value of these functions without finding the roots themselves.

Common Examples:
To find \(\alpha^2 + \beta^2 + \gamma^2\), use the identity:
\((\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
Rearranging this gives:
\(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)

Don't worry if this seems tricky! Just remember that your goal is always to rewrite the expression using only the "building blocks" we learned in Section 1 (the sum, the sum of pairs, etc.).

Common Mistake to Avoid: Thinking that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2\). It’s not! You must subtract the "extra" middle term: \(2\alpha\beta\).

3. Root Transformations (Using Substitutions)

Sometimes, a problem will give you an equation with roots \(\alpha, \beta, \gamma\) and ask you to find a new equation with roots that are related, such as \(2\alpha, 2\beta, 2\gamma\) or \(\alpha^2, \beta^2, \gamma^2\).

Instead of finding the roots, we use substitution. This is like a "mathematical makeover" for the equation.

Step-by-Step Process:

1. Let \(y\) be the new root you want (e.g., if the new roots are squares, let \(y = x^2\)).
2. Rearrange this to make \(x\) the subject (e.g., \(x = \sqrt{y}\)).
3. Substitute this expression for \(x\) back into the original equation.
4. Simplify the equation so it looks like a standard polynomial again (no square roots or fractions in the denominators).

Example: Find an equation with roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\) for the original equation \(x^3 + 2x^2 - 5x + 6 = 0\).
Step 1: Let \(y = \frac{1}{x}\).
Step 2: Therefore, \(x = \frac{1}{y}\).
Step 3: Replace \(x\): \((\frac{1}{y})^3 + 2(\frac{1}{y})^2 - 5(\frac{1}{y}) + 6 = 0\).
Step 4: Multiply everything by \(y^3\) to clear the fractions: \(1 + 2y - 5y^2 + 6y^3 = 0\).
Rearranged: \(6y^3 - 5y^2 + 2y + 1 = 0\).

Analogy: Think of substitution like a translator. If the original roots speak "X", and you need them to speak "Y", you use the substitution rule to translate the entire sentence (the equation) so that it works for the new language.

Key Takeaway: Substitution is the fastest way to create a new equation. Just set \(y\) equal to the new root, solve for \(x\), and plug it back in.

4. Solving for Unknown Coefficients

If the exam tells you something specific about the roots (e.g., "one root is double the other" or "the roots are in arithmetic progression"), you can use the relations from Section 1 to find missing numbers in the equation.

Example Trick: If roots are in arithmetic progression, you can call them \(k-d\), \(k\), and \(k+d\). When you add them together (\(\sum \alpha\)), the \(d\)'s cancel out, making it very easy to find the value of \(k\)!

Summary Checklist for Success:

• Do I remember the sign pattern (\(-, +, -, +\))?
• Did I divide the coefficients by \(a\)?
• When using substitution, did I simplify the final equation fully?
• Have I checked for "missing" terms? (e.g., if there is no \(x^2\) term, that coefficient is 0).

Final Encouragement: Polynomial roots might look intimidating with all those Greek letters, but they follow very strict rules. Master the "Sum and Product" patterns, and you've already won half the battle!