Solution to Quadratic Equation

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Theorem

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

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


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:

\(\displaystyle 4 a^2 x^2 + 4 a b x + 4 a c\) \(=\) \(\displaystyle 0\) $\quad$ multiplying through by $4 a$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({2 a x + b}\right)^2 - b^2 + 4 a c\) \(=\) \(\displaystyle 0\) $\quad$ Completing the Square $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({2 a x + b}\right)^2\) \(=\) \(\displaystyle b^2 - 4 a c\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \frac {-b \pm \sqrt {b^2 - 4 a c} }{2a}\) $\quad$ $\quad$

$\blacksquare$


Also known as

This result is often referred to as the quadratic formula.


Sources