#A05 · Algebra & Number Theory · Lv.2View in 3D Mind-Map

Vieta's Formulas (Roots & Coefficients)

François Viète · 1615 CE

\alpha + \beta = -\frac{b}{a},\quad \alpha\beta = \frac{c}{a}

a(x - \alpha)(x - \beta) = ax^2 + bx + c

Relates polynomial roots directly to coefficients without solving for the roots individually; directly generalizes to matrix trace (sum of eigenvalues) and determinant (product of eigenvalues).

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Root α

Root β

x^2 - (2.0)x + (-8.0) = 0 \iff \alpha+\beta=2.0,\; \alpha\beta=-8.0

두 근의 합 α+β = -b/a

2.000

두 근의 곱 αβ = c/a

-8.000

대칭식 α²+β²

20.000

Prerequisite Path (5 STEP)

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Click steps in order to follow the 3D neon path from basics to target.

STEP 1#A01Linear Equation & Balance Axiom

ax + b = 0 \implies x = -\frac{b}{a}

STEP 2#A03Laws of Exponents & Radicals

a^m \cdot a^n = a^{m+n}

STEP 3#A02Polynomial Expansion & Factoring

(a+b)^2 = a^2 + 2ab + b^2

STEP 4#A04Quadratic Formula & Discriminant

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

STEP 5#A05Vieta's Formulas (Roots & Coefficients)

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

🚀 Next Unlocks (Derived Formulas) (1)

#L08Eigenvalues & Eigenvectors (Av = λv)

Geometric & Intuitive Insight

a(x - α)(x - β) 를 전개하면 ax² - a(α+β)x + aαβ 가 되므로, 원래 식 ax² + bx + c 와 계수를 비교하면 b = -a(α+β), c = aαβ 가 자명하게 성립합니다.

Step-by-Step Derivation

Step 1.1단계: 두 근을 α, β라 하면 이차방정식은 a(x - α)(x - β) = 0 으로 인수분해됩니다.
Step 2.2단계: 좌변을 전개하면 ax² - a(α + β)x + aαβ = 0 이 됩니다.
Step 3.3단계: ax² + bx + c = 0 과 계수를 비교하면 α + β = -b/a, αβ = c/a 를 얻습니다.

AI & Coding Real-World Use

Control theory pole stability checks and matrix eigenvalue trace/determinant verification

#Matrix Trace & Det#Control Theory Poles#Symmetric Polynomials

const sumRoots = -b / a; const prodRoots = c / a;

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