<s>[INST] Help with basic high school math. What happens to $j$? I know my math is very rusty, actually, its always been that way. but I need help with this. The question below has me stumped. I've tried to show the steps I went through to get the answer. Please tel me where I made the mistake.

If x=a and x=b are two roots of a quadratic equation then (x-a)(x-b) = 0 gives the quadratic equation.
That is $(x - a)(x - b) = x^2 - (a + b)x + ab = 0$.
Here, the two roots are $x= -2 + j\sqrt5$ and $x = -2 - j\sqrt5$ so that
$(x – [-2 + j\sqrt5])(x – [-2 - j\sqrt5]) = 0$
That is $x^2 - x[-2 + j\sqrt5 - 2 - j\sqrt5] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
I understand that
$x^2 - x[-2 + j\sqrt5 - 2 - j\sqrt5] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
$x^2 - x[-2 - 2] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$, and
$x^2 - x[-4] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$, so
$x^2 + 4x + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
if we separate out the last term for simplicity:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + (+j\sqrt5)(-j\sqrt5)$
$= 4 + (2j\sqrt5) - (2j\sqrt5) +(-j\sqrt5)^2$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) +(-j^2)(-\sqrt5)^2$
$ = 4 +j^2 5$
Putting this last term back into the main equation results in:
$x^2 + 4x + (4+j^2 5) = 0$
In the book (Advanced Engineering Mathematics)* this equation works out to
$x^2 + 4x + 9 = 0$
What I don’t understand is what happened to $j^2$ How does it just magically disapear?
*If you use the Amazon "Look inside" feature you can see it on page 4.
 [/INST] Here are your steps with mistakes highlighted in red:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + (+j\sqrt5)(-j\sqrt5)$
$= 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j\sqrt5)^2}$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j^2)(-\sqrt5)^2}$
$ = 4 + \color{red}{j^2 5}$
Here is fixed, with highlighting in red:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + \color{blue}{(+j\sqrt5)(-j\sqrt5)}$
$= 4 + (2j\sqrt5) - (2j\sqrt5) +\color{red}{(-1)\times j^2\times (\sqrt5)^2}$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) +\color{red}{(-1)\times (-1)\times 5}$
$ = 4 +\color{red}{5} = 9$
Note that
$$\color{blue}{(+j\sqrt5)(-j\sqrt5)} =
\color{red}{(-1)\times j^2\times (\sqrt5)^2}$$
and
$$j^2 = -1$$
</s>