Solution to Quadratic Equation

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Theorem

The quadratic equation of the form $a x^2 + b x + c = 0$ has solutions:

$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$


Real Coefficients

Let $a, b, c \in \R$.

The quadratic equation $a x^2 + b x + c = 0$ has:

Two real solutions if $b^2 - 4 a c > 0$
One real solution if $b^2 - 4 a c = 0$
Two complex solutions if $b^2 - 4 a c < 0$, and those two solutions are complex conjugates.


Proof

Let $a x^2 + b x + c = 0$. Then:

\(\ds 4 a^2 x^2 + 4 a b x + 4 a c\) \(=\) \(\ds 0\) multiplying through by $4 a$
\(\ds \leadsto \ \ \) \(\ds \paren {2 a x + b}^2 - b^2 + 4 a c\) \(=\) \(\ds 0\) Completing the Square
\(\ds \leadsto \ \ \) \(\ds \paren {2 a x + b}^2\) \(=\) \(\ds b^2 - 4 a c\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}\)

$\blacksquare$


Also known as

This result is often referred to as the quadratic formula.


Sources

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