Viète's Formulas/Examples/Cubic

Example of Use of Viète's Formulas

Consider the cubic equation:

$x^3 + a_1 x^2 + a_2 x + a_3 = 0$

Let its roots be denoted $x_1$, $x_2$ and $x_3$.

Then:

\(\ds x_1 + x_2 + x_3\)

\(=\)

\(\ds -a_1\)

\(\ds x_1 x_2 + x_2 x_3 + x_3 x_1\)

\(=\)

\(\ds a_2\)

\(\ds x_1 x_2 x_3\)

\(=\)

\(\ds -a_3\)

Proof

A specific instance of Viète's Formulas for $n = 3$.

$\blacksquare$

Sources

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Cubic Equations: $3.8.2$
• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 9$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.5$
• 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.5.$