anonymous
  • anonymous
Find the square roots of -3+4i
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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EarthCitizen
  • EarthCitizen
square roots ?
anonymous
  • anonymous
5.73205081 i according to google :/
EarthCitizen
  • EarthCitizen
i do know if you convert to polar form, r = 5(radius),theta =126.87°(angle)

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anonymous
  • anonymous
I haven't learned polar form yet... It's using the (a^2-b^2)+2abi method.
anonymous
  • anonymous
How do you factor a^4+3a^2+4=0?
Callisto
  • Callisto
i think you cannot factor it ?! (not include using the quadratic formula)
EarthCitizen
  • EarthCitizen
with that you get four roots then
anonymous
  • anonymous
When it's a^4-3a^2-4=0, it factors as (a^2-4)(a^2+1)..?
EarthCitizen
  • EarthCitizen
@order (a^2-a+2) (a^2+a+2) = 0, is a = -3+j4 ?
anonymous
  • anonymous
Not sure... does this have to do with my question?
EarthCitizen
  • EarthCitizen
yep, yep
anonymous
  • anonymous
(1 + 2i)^2 = -3+4i
anonymous
  • anonymous
|dw:1333533962294:dw| solve that system to get a and b...
anonymous
  • anonymous
you should be solving: a^4 + 3a^2 - 4 = 0. i noticed you have something different above...
anonymous
  • anonymous
\[ a^4 + 3a^2 - 4 = 0. \] \[ (a^2 +4) (a^2-1) =0 \] \[ a =\pm 2i \text{ and } a =\pm 1 \] Since \[ a\] is real so \[ a =\pm 1 \] is the only acceptable solution. Since ab =2 then the square roots are -1 -2 i and 1+2 i
anonymous
  • anonymous
why -4?
anonymous
  • anonymous
|dw:1333536526613:dw|
anonymous
  • anonymous
sorry, i can't use latex like these pros...

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