Sum product roots

Connecting Roots to Coefficients

If a quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$) has roots $\alpha$ and $\beta$, we can write it in factored form as $a(x - \alpha)(x - \beta) = 0$.

Expanding this factored expression:
$a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] = 0 \implies ax^2 - a(\alpha + \beta)x + a\alpha\beta = 0$

Comparing coefficients with standard form $ax^2 + bx + c = 0$:

• Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$
• Product of roots: $\alpha\beta = \dfrac{c}{a}$

This connects the roots directly to the coefficients without needing to solve the equation.

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Reconstructing Quadratics & Symmetric Identities

Reconstructing an Equation

Dividing $ax^2 + bx + c = 0$ by $a$ gives $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$. Substituting our root relationships:
$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$

Key Symmetric Identities

Since $\alpha$ and $\beta$ appear symmetrically, we can express various algebraic combinations using only $S = (\alpha + \beta)$ and $P = (\alpha\beta)$:

• Sum of squares: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
• Difference squared: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$
• Sum of cubes: $\alpha^3 + \beta^3 = (\alpha + \beta)\left[(\alpha + \beta)^2 - 3\alpha\beta\right]$

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Worked Example: Evaluating Expressions

Problem: Let $\alpha$ and $\beta$ be the roots of $2x^2 - 5x - 1 = 0$. Without solving for $\alpha$ and $\beta$, evaluate:

1. $\alpha + \beta$ and $\alpha\beta$
2. $\alpha^2 + \beta^2$
3. $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$

Step 1: Identify coefficients and find sum & product

• $a = 2, \quad b = -5, \quad c = -1$
• $\alpha + \beta = -\dfrac{b}{a} = -\dfrac{-5}{2} = \dfrac{5}{2}$
• $\alpha\beta = \dfrac{c}{a} = -\dfrac{1}{2}$

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Worked Example: Step-by-Step Solution

Step 2: Evaluate $\alpha^2 + \beta^2$

Apply the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{25}{4} + 1 = \frac{29}{4}$

Step 3: Evaluate $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$

Combine into a single fraction:
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{5}{2}}{-\frac{1}{2}} = -5$

Common Pitfall: Don't forget the negative sign in $\alpha + \beta = -\frac{b}{a}$. Here, $b = -5$, so $-\frac{b}{a} = +\frac{5}{2}$.