Core Concept
MemorizeVieta's Formulas — The Foundation
For the quadratic equation \(ax^2 + bx + c = 0\) with two roots \(\alpha\) and \(\beta\):
Vieta's Formulas
\(\alpha + \beta = -\dfrac{b}{a}\)
\(\alpha \cdot \beta = \dfrac{c}{a}\)
\(\alpha \cdot \beta = \dfrac{c}{a}\)
Sum of Roots
\(\alpha+\beta = -\dfrac{b}{a}\)
Product of Roots
\(\alpha\beta = \dfrac{c}{a}\)
Reverse: Build Equation
\(x^2 - Sx + P = 0\)
Key Derived Formula
\(\alpha^2+\beta^2 = S^2-2P\)
Memory Trick: Sum = \(-b/a\) (negative sign!), Product = \(c/a\) (positive). For the monic case \(x^2+bx+c=0\): Sum\(= -b\), Product\(= c\).
Essential Derived Expressions
Let \(S = \alpha+\beta\) and \(P = \alpha\beta\).
\(\alpha^2 + \beta^2 = S^2 - 2P\)
\((\alpha - \beta)^2 = S^2 - 4P\)
\(\alpha^3 + \beta^3 = S^3 - 3PS\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{S}{P}\) (when \(P \neq 0\))
\(\dfrac{1}{\alpha\beta} = \dfrac{1}{P}\)
\(\alpha^2 + \beta^2 = S^2 - 2P\)
\((\alpha - \beta)^2 = S^2 - 4P\)
\(\alpha^3 + \beta^3 = S^3 - 3PS\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{S}{P}\) (when \(P \neq 0\))
\(\dfrac{1}{\alpha\beta} = \dfrac{1}{P}\)
Worked Examples
Study First
Example 1
Direct Application
The roots of \(2x^2 - 5x + 3 = 0\) are \(\alpha\) and \(\beta\). Find \(\alpha + \beta\) and \(\alpha\beta\).
Here \(a=2,\; b=-5,\; c=3\).
\(\alpha + \beta = -\dfrac{b}{a} = -\dfrac{-5}{2} = \dfrac{5}{2}\)
\(\alpha\beta = \dfrac{c}{a} = \dfrac{3}{2}\)
Answer: \(\alpha+\beta = \dfrac{5}{2},\quad \alpha\beta = \dfrac{3}{2}\)
Example 2
Derived Expression
Roots of \(x^2 - 4x + 1 = 0\) are \(\alpha,\beta\). Find \(\alpha^2 + \beta^2\).
\(S = \alpha+\beta = 4,\quad P = \alpha\beta = 1\)
\(\alpha^2+\beta^2 = S^2 - 2P = 16 - 2 = 14\)
Answer: \(\alpha^2+\beta^2 = 14\)
Example 3
Reverse — Build the Equation
Form a monic quadratic with roots \(\alpha = 3\) and \(\beta = -5\).
\(S = 3+(-5) = -2,\quad P = 3\cdot(-5) = -15\)
Equation: \(x^2 - Sx + P = 0 \;\Rightarrow\; x^2 + 2x - 15 = 0\)
Answer: \(x^2 + 2x - 15 = 0\)
Practice Problems
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