1.1 Roots of Polynomial Equations

1.1 Quadratic Equations 🔑

Relations Between Roots and Coefficients

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we can rewrite it in the form:

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

Comparing this to the factored form $(x - \alpha)(x - \beta) = 0$, which expands to:

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

We obtain the fundamental relationships:

Sum of roots:

$\alpha + \beta = -\frac{b}{a}$

Product of roots:

$\alpha\beta = \frac{c}{a}$

General Quadratic Form

Any quadratic equation with roots $\alpha$ and $\beta$ can be written as:

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

or equivalently:

$x^2 - \left(\sum \alpha\right)x + \sum \alpha\beta = 0$

Sum of Powers of Roots

We can find expressions for sums of powers of roots using algebraic manipulation.

Sum of squares:

Starting with $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, we rearrange to get:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

In summation notation:

$\sum \alpha^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta$

Sum of cubes:

Using the binomial expansion:

$(\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3$

Factoring the middle terms:

$(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)$

Rearranging:

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

In summation notation:

$\sum \alpha^3 = \left(\sum \alpha\right)^3 - 3\sum \alpha\beta \sum \alpha$

Recurrence Relations ⚡

For a quadratic equation $ax^2 + bx + c = 0$, we can define:

$S_n = \alpha^n + \beta^n$

Since $\alpha$ and $\beta$ are roots, they satisfy:

Multiplying the first equation by $\alpha^n$ and the second by $\beta^n$, then adding:

$a(\alpha^{n+2} + \beta^{n+2}) + b(\alpha^{n+1} + \beta^{n+1}) + c(\alpha^n + \beta^n) = 0$

This gives the recurrence formula:

$aS_{n+2} + bS_{n+1} + cS_n = 0$

This formula allows us to calculate $S_n$ for any positive integer $n$ once we know $S_0$ and $S_1$.

Worked Example 1: Finding Unknown Coefficients

Problem: The quadratic equation $x^2 - 2px + p = 0$ is such that one root is three times the value of the other root. Find $p$.

Solution:

Let the roots be $\alpha$ and $3\alpha$.

Sum of roots:

$\alpha + 3\alpha = 2p$

$4\alpha = 2p$

$p = 2\alpha \quad \implies \quad \alpha = \frac{p}{2}$

Product of roots:

$\alpha \times 3\alpha = p$

$3\alpha^2 = p$

Equating the two expressions:

Substitute $\alpha = \frac{p}{2}$ into $3\alpha^2 = p$:

$3\left(\frac{p}{2}\right)^2 = p$

$3 \cdot \frac{p^2}{4} = p$

$\frac{3p^2}{4} = p$

$3p^2 = 4p$

$3p^2 - 4p = 0$

$p(3p - 4) = 0$

Since we need $p > 0$ for real distinct roots:

$p = \frac{4}{3}$

Worked Example 2: Using Recurrence Relations

Problem: Given that $2x^2 + 3x - 2 = 0$ has roots $\alpha, \beta$, find the values of $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.

Solution:

From the equation $2x^2 + 3x - 2 = 0$, we have $a = 2$, $b = 3$, $c = -2$.

Finding $S_1$:

$S_1 = \alpha + \beta = -\frac{b}{a} = -\frac{3}{2}$

Finding $S_2 = \alpha^2 + \beta^2$:

The recurrence relation is $2S_{n+2} + 3S_{n+1} - 2S_n = 0$.

For $n = 0$:

$2S_2 + 3S_1 - 2S_0 = 0$

$2S_2 + 3\left(-\frac{3}{2}\right) - 2(2) = 0$

$2S_2 - \frac{9}{2} - 4 = 0$

$2S_2 = \frac{9}{2} + 4 = \frac{9}{2} + \frac{8}{2} = \frac{17}{2}$

$S_2 = \frac{17}{4}$

Finding $S_3 = \alpha^3 + \beta^3$:

Multiply the original equation by $x$:

$2x^3 + 3x^2 - 2x = 0$

This gives us the recurrence:

$2S_3 + 3S_2 - 2S_1 = 0$

$2S_3 + 3\left(\frac{17}{4}\right) - 2\left(-\frac{3}{2}\right) = 0$

$2S_3 + \frac{51}{4} + 3 = 0$

$2S_3 + \frac{51}{4} + \frac{12}{4} = 0$

$2S_3 + \frac{63}{4} = 0$

$S_3 = -\frac{63}{8}$