4
$\begingroup$

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.

  • 0
    @Gineer: [This answer](http://math.stackexchange.com/questions/123652/multiplying-exponents-with-variables-inside) of Arturo Magidin might help you get the signs correctly.2012-04-03

1 Answers 1

5

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

$j^2 = -1$