Solution to Quadratic Equation/Real Coefficients

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Theorem

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

From Solution to Quadratic Equation:

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


If the discriminant $b^2 - 4 a c > 0$ then $\sqrt {b^2 - 4 a c}$ has two values and the result follows.


If the discriminant $b^2 - 4 a c = 0$ then $\sqrt {b^2 - 4 a c} = 0$ and $x = \dfrac {-b} {2 a}$.


If the discriminant $b^2 - 4 a c < 0$, then it can be written as:

$b^2 - 4 a c = \paren {-1} \size {b^2 - 4 a c}$

Thus:

$\sqrt {b^2 - 4 a c} = \pm i \sqrt {\size {b^2 - 4 a c} }$

and the two solutions are:

$x = \dfrac {-b} {2 a} + i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}, x = \dfrac {-b} {2 a} - i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}$

and once again the result follows.

$\blacksquare$


Sources

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