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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

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you cannot solve $z^2 - 6z + 25$ but you can solve $z^2 - 6z + 25 = 0$. –  anon Oct 23 '10 at 13:47
    
thats what I meant, I edited it. –  gideon Oct 23 '10 at 13:49
    
Well, it could have also been presented as "factor quadratic so-and-so into linear factors". –  J. M. Oct 23 '10 at 13:52
    
+1 for showing what you had tried –  Ross Millikan Oct 23 '10 at 16:47

4 Answers 4

up vote 6 down vote accepted

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:

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

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

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

$$z - 3 = \pm 4i$$

$$z = 3 \pm 4i$$

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oooo ahhh so we're solving for z! It sucks having to figure out math on my own, I think I would've failed a third time if it were not for forums like this (I used to post a lot on another math forum) I just hope I get a D this time!!! I'll just be decently happy... –  gideon Oct 23 '10 at 13:55
3  
@giddy: That's right... Don't despair, just keep tackling these sorts of problems and at some point it will "click", and you'll no longer find it a pain! :) –  Noldorin Oct 23 '10 at 14:57

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.

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but how does -16 get in there? The square root of (z-3)^2 + 16 should be z -3 + 4 right? –  gideon Oct 23 '10 at 13:46
2  
Nope. The sum of two squares $x^2+y^2$ factorizes differently from the perfect square trinomial $x^2+2xy+y^2=(x+y)^2$. –  J. M. Oct 23 '10 at 13:48
1  
Giddy, be very careful when taking the square root of a sum or difference of squares. The error J. M. pointed out is a "dangerous" one-- take your time understanding why what you just said was wrong... it will pay off in your future studies. –  a little don Oct 24 '10 at 0:18
    
haha ok yea I see... when I wrote it down in my book and then got it man I can be silly sometime! –  gideon Oct 24 '10 at 16:00

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 $3-4i$.

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this does seem nice and easy since I forget how to complete the square. –  gideon Oct 24 '10 at 15:45

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 $

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