Sum of Roots of Quadratic Equation

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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


Sources