Using Method 2 for quadratics, we have
\begin{align}
x^2 + 4 &= 0 \\
x^2 &= -4 \\
\sqrt{x^2} &= \pm \sqrt{-4}
\end{align}
To evaluate $\sqrt{-4}$ we need to find a number $a$ where $a^2 = -4$, a negative number.
This is impossible. If $a$ is positive, $a^2$ is positive.
If $a^2$ is negative, then $a^2$ is still positive.
Therefore we need a number $a$ where $a^2$ can be negative!
complex number
A complex number has the form \[a + bi\] where $a, b$ are real numbers and $i^2 = -1$. Also, $i = \sqrt{-1}$.
$a$ is called the real part, while $b$ is called the imaginary part.
In our introductory example, we still need to take the square root of $-4$, a negative number.
Here's how.
Square Roots of Negative Numbers
If $-r$ is negative, the principal square root of $-r$ is \[\sqrt{-r} = \sqrt{(-1)r} = \sqrt{-1}\sqrt{r} = i\sqrt{r}\]
If variables are involved, the two square roots of $-r$ are $i\sqrt{r}$ and $-i\sqrt{r}$.
Simplify $\sqrt{-16}$.
Solution: We have \[\sqrt{-16} = i\sqrt{16} = i\cdot 4 = \boxed{4i}\]
Simplify $\sqrt{-3}$.
Simplify $\sqrt{-2}\sqrt{-3}$.
Complex Solutions of Quadratic Equations
Recall that a quadratic equation $ax^2 + bx + c = 0$ has solutions \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
$b^2 - 4ac$ in the square root determines what type of solutions a quadratic could have.
For the quadratic equation $ax^2 + bx + c = 0$, the quantity $b^2 - 4ac$ is called the discriminant.
If $b^2 - 4ac > 0$, there are two distinct (different) real solutions.
If $b^2 - 4ac = 0$, there is one real solution.
If $b^2 - 4ac < 0$, there are two distinct complex solutions.
Solve $x^2 + 4 = 0$.
We have
\begin{align}
x^2 + 4 &= 0 \\
x^2 &= -4 \\
\sqrt{x^2} &= \pm \sqrt{-4} \\
x &= \pm i\sqrt{4} \\
&\boxed{x = \pm 2i}
\end{align}
Solve $x^2 + 4x + 5 = 0$.
Using the quadratic formula with $a = 1, b = 4, c = 5$:
\begin{align}
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
&= \dfrac{-4 \pm\sqrt{4^2 - 4\cdot 1 \cdot 5}}{2\cdot 1} \\
&= \dfrac{-4 \pm\sqrt{16 - 20}}{2} \\
&= \dfrac{-4 \pm \sqrt{-4}}{2} \\
&= \dfrac{-4 \pm i\sqrt{4}}{2} \\
&= \dfrac{-4 \pm 2i}{2} \\
&= \dfrac{2(-2\pm i)}{2} \\
&= \boxed{-2 \pm i}
\end{align}