One solution to a quadratic equation is x equals start fraction negative nine plus 21 i over three end fraction full stop What is the solution in simplified standard form, x = a + bi, if a and b are real numbers? Help please

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One solution to a quadratic equation is x equals start fraction negative nine plus 21 i over three end fraction full stop What is the solution in simplified standard form, x = a + bi, if a and b are real numbers? Help please

Mathematics
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\[\frac{-9+21i}{3}=\frac{-9}{3}+\frac{21i}{3}=-3+7i\]
Thank you bunches. :) Can you help me with 2 more?
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The complex solution to a quadratic equation is x equals start fraction negative four plus or minus square root of negative 80 end square root over four end fraction full stop Write this solution in standard form, x = a ± bi, where a and b are real numbers.
\[x=\frac{-4\pm\sqrt{-80}}{4}\]?
yes
ok we look for the largest perfect square that is a factor of 80 i get 16 because \(80=16\times 5\)
okay that makes sense
therefore \[\sqrt{-80}=\sqrt{16\times 5\times (-1)}=4\sqrt{5}i\]
woah you lost me D: lol
i guess i skipped a step \[\sqrt{16\times 5\times (-1)}=\sqrt{16}\sqrt{5}\sqrt{-1}=4\sqrt{5}i\]
since \(\sqrt{16}=4\) and \(\sqrt{-1}\) gets written as \(i\)
that is why we were looking for the largest perfect square that is a factor of 80
let me know when we can continue, it is not over yet
okay that makes sense
that was the hard part though now we have \[\frac{-4\pm4\sqrt{5}i}{4}\]divide each part of the numerator by \(4\) and you are done
this is what you have to do for all of them write the radical in simplest radical form for example since \(25\times2=50\) you would have \[\sqrt{-50}=5\sqrt{2}i\]
The number root of order five of ninety one to the power of four end power can be written as ninety one to the power of start fraction cap a over cap b end fraction end power. What is the value of A?

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