# 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.

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The last line in the sequence of equations after "if we seperate put the last term for simplicity" should be $4-j^25$, as $(-j^2)(-\sqrt{5})^2=-j^25$. This gives us $4-5j^2=4-(-1)\cdot 5=9$, because $j=\sqrt{-1}$. – Alex Becker Apr 3 '12 at 21:12
You have to be careful with signs: $(-j)(+j) = -j^2$ is not the same as $(-j)^2$. – Dejan Govc Apr 3 '12 at 21:17
As the son of an electrical engineer, I learned to use the letter $j$ for $\sqrt{-1}$ before I learned to use $i$. Here's something that's happened to me several times with undergraduates. I'm doing some routine calculation in which $i^2$ appears, and a student show claims to be well aware of the fact that $i=\sqrt{-1}$ does not know what $i^2$ is. I ask if they know what $\sqrt{83}\cdot\sqrt{83}$ is, and they say: not without a calculator. I ask: Do you know what square roots are? They say "yes". Then I say $\sqrt{83}\cdot\sqrt{83}$ is $83$. Then I'm asked how I got that. – Michael Hardy Apr 3 '12 at 21:19
@MichaelHardy $*$ shudder $*$ – anon Apr 3 '12 at 21:27
@anon You think that's bad? As the son of an Econ professor, I hear way worse. The one that takes the cake is a student asking for a calculator on an exam to simplify $\frac{120}{1}$. – Alex Becker Apr 3 '12 at 21:36

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

$$j^2 = -1$$

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