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Use the quadratic formula to find the complex roots (Leave answer in simplest a+bi form) : 3x²-2x+1=0
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deja vu that strange feeling you get when you think you've been there before
didn't we just do this and get \[\frac{1\pm\sqrt{2}i}{2}\]
Sorry, I had the question wrong
if you want to make it look like \(a+bi\) split it up as \[\frac{1}{2}+\frac{\sqrt{2}}{2}i\]
is that the simplest a+bi form?
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oh
yes
but, I was curious why 2 isnt multiplied by 3 on the denominator
\[\frac{ 2\pm \sqrt{-8} }{ 2(3) }\]
a=3 b=-2 c=1 x=[-b +- sqrt(b^2 -4*a*c)] / 2*a x= [--2 +- sqrt(4 - 4*3*1)]/ 6 x = [2 +- sqrt(4 -12)]/6 x = [2 +- sqrt(8)] / 6
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can I convert to a+bi form
|dw:1425863896183:dw|
|dw:1425863918782:dw|
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