Sum of Roots of Quadratic Equation

Theorem

Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.

Let $\alpha$ and $\beta$ be the roots of $P$.

Then:

$\alpha + \beta = -\dfrac b a$

Proof

\(\ds \alpha\)

\(=\)

\(\ds \frac {-b + \sqrt {b^2 - 4 a c} } {2 a}\)

Solution to Quadratic Equation

\(\ds \beta\)

\(=\)

\(\ds \frac {-b - \sqrt {b^2 - 4 a c} } {2 a}\)

Without loss of generality, selecting $\alpha$ and $\beta$ as such

\(\ds \leadsto \ \ \)

\(\ds \alpha + \beta\)

\(=\)

\(\ds \frac {-b - \sqrt {b^2 - 4 a c} - b + \sqrt {b^2 - 4 a c} } {2 a}\)

\(\ds \)

\(=\)

\(\ds \frac {-2 b} {2 a}\)

\(\ds \)

\(=\)

\(\ds -\frac b a\)

$\blacksquare$

Also see

• Product of Roots of Quadratic Equation
• Viète's Formulas

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 9$: Solutions of Algebraic Equations: $9.2$: Quadratic Equation
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): quadratic equation