Pre-Calculus: Complex Zeros http://prntscr.com/aaigd9 So I solved with the quadratic equation and got these results but they're wrong ?? And I checked on multiple sites to make sure I did everything right e_O
the form is a+bi did you try this way: -1/12 - i \sqrt(143)/12 , .. ??
yes, but it was wrong so I put i .... wait
I think I made a stupid mistake...
also the first root has a sign mistake
what do you mean?
no no, they both are correct..!
oh... o_o
oh yess @hartnn i had a question on another thing but nobody responded... it is closed question now but can i tag you in it?
yes* oops...
http://www.wolframalpha.com/input/?i=x%5E2%2Bx%2F6%2B1+%3D0 yeah, i saw it, and helping that user already, thanks for tagging! :)
no not that... i didn't tag you in my own question... it was the one before this one but jdoe didn't respond so i closed it lol
lol, can you tag me there again? :P
ok I will tag you... brb
we factored out x^2, yes? try `x^2, (x+9i)(x-9i)`
without the comma :P
OHHH right the multiplicity D:
ok... I think I get it now...
no... I got the second one wrong still @hartnn sorry http://prntscr.com/aaipol
It asks you factor it completely so you need to use sum of squares and difference of squares
\[(x+i 5)(x- i 5)(x+5)(x-5)\]
Actually your answer doesn't look right it should be \[(x^2+25)^2 \] yeah?
really? o: @Astrophysics
Yes, lets confirm: http://www.wolframalpha.com/input/?i=factor+x%5E4%2B50x%5E2%2B625
\[(x^2+25)^2 \neq (x+5)(x-5)(x-5 i ) (x+5 i )\]
It's sum of squares twice
Oh... I don't even know on here anymore lol. I fixed it at school with the help of my teacher so :T But! MEDALS TO Y'ALL thanks for helping ~ ♥
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