I've been working on this equation, here's what I have so far:

$\displaystyle z^2-4z+26=0$

$\displaystyle z^2-4z+4=30$

$\displaystyle (z-2)^2=30$

$\displaystyle z-2= +- Square root of 30$

...

And here I'm lost. Please help.

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- Jan 12th 2012, 01:03 PMHellBunnyQuadratic Equation
I've been working on this equation, here's what I have so far:

$\displaystyle z^2-4z+26=0$

$\displaystyle z^2-4z+4=30$

$\displaystyle (z-2)^2=30$

$\displaystyle z-2= +- Square root of 30$

...

And here I'm lost. Please help. - Jan 12th 2012, 01:08 PMPlatoRe: Quadratic Equation
- Jan 12th 2012, 01:10 PMe^(i*pi)Re: Quadratic Equation
- Jan 12th 2012, 01:12 PMHellBunnyRe: Quadratic Equation
Okay ( to first poster )

(Second poster)

$\displaystyle [1/2(-4)]^2$

Is how I got the input. - Jan 12th 2012, 01:15 PMSironRe: Quadratic Equation
This is not correct. It has to be:

$\displaystyle z^2-4z+26=0$

$\displaystyle \Leftrigharrox (z^2-4z+4)+22=0$

$\displaystyle \Leftrightarrow (z-2)^2=-22$

$\displaystyle \Lefrightarrow z-2= \pm \sqrt{-22}$

$\displaystyle \Leftrightarrow z=\pm \sqrt{-22} + 2$

Now, what's $\displaystyle \sqrt{-22}$?

Note that $\displaystyle D<0$ therefore there're only complex roots, like we obtained. - Jan 12th 2012, 01:16 PMe^(i*pi)Re: Quadratic Equation
I still don't follow what you did to the value of "c".

If you took 22 from both sides (which is what the LHS suggests) then you'd have -22 on the right (giving complex solutions)

If you added 30 to both sides as implied then you'd have +56 on the left.

Edit: too slow - Jan 12th 2012, 01:23 PMHellBunnyRe: Quadratic Equation
It's fine ^. Thanks, that's the answer! Thank you all again for helping.