6
$\begingroup$

I need to solve this :$ z^2 -6z + 25 = 0$

My book says 'complete the square' so :
1.$ (z - 6/2)^2 -36/4 + 25 $
2.$ (z - 3)^2 -9 + 25 $
3.$ (z - 3)^2 + 16 $

Now how exactly does the above turn into this:
$ 3\pm 4\imath $

Thanks so Much
Gideon

  • 0
    Well, it could have also bee$n$ presented as "factor quadratic so-and-so into linear factors".2010-10-23

4 Answers 4

6

I think your confusion here stems from the fact you're working with an expression, not an equation. You can only solve equations; it makes no sense to 'solve' an expression!

Edit: You seem to have just edited your post so the first line is an equation. Best to keep it in equation form throughout however.

I'm presuming you have:

$z^2 - 6z + 25 = 0$

Then, completing the square:

$(z - 6/2)^2 -36/4 + 25 = 0$

$(z - 3)^2 - 9 + 25 = 0$

$(z - 3)^2 = -16$

$z - 3 = \pm 4i$

$z = 3 \pm 4i$

  • 3
    @giddy: Th$a$t's right... Don't despair, just keep tackling these sorts of pro$b$lems and at some point it will "click", and you'll no longer find it a pain! :)2010-10-23
3

The square roots of -16 are $4i$ and $-4i$.


As a further hint to anybody who might encounter this sort of thing in the future, a polynomial with complex conjugate roots $\alpha$ and $\alpha^\ast$ will be of the form

$x^2-2(\Re\alpha)x+|\alpha|^2$

Note that $3^2+4^2=5^2=25$ and that $6=2\times 3$. Thus, once you verify that your quadratic has a negative discriminant, you can almost instantly write down the complex conjugate roots.

  • 0
    haha ok yea I see... when I wrote it down in my book and then _got it_ man I can be silly sometime!2010-10-24
1

Noldorin's answer pretty much covers what I'd say, but I would note that you can proceed in working with the expression until it is factored into linear terms and read off the zeros (the solutions to the equation $(z - 3)^2 + 16=0$) from there.

Continuing from $(z - 3)^2 + 16$: $\begin{align} (z - 3)^2 + 16 &= (z - 3)^2 - (-16) \\\\ &= (z - 3)^2 - (4i)^2 \\\\ &= ((z-3)-4i)((z-3)+4i) \\\\ &= (z-(3+4i))(z-(3-4i)) \end{align}$ So, the zeros of the expression are $3+4i$ and $3-4i$.

  • 0
    this does seem nice and easy since I forget how to complete the square.2010-10-24
0

You can also use the quadratic formula to compute the result elegantly, it goes like this :

$z^2 - 6z + 25 = 0 $

$ z = \frac { 6 \pm \sqrt{36 -100} } {2} $

$ = \frac{ 6 \pm 8i}{2} $

$= 3 \pm 4i $