Question

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

Answer 1

\[\frac{-9+21i}{3}=\frac{-9}{3}+\frac{21i}{3}=-3+7i\]

Answer 2

Thank you bunches. :) Can you help me with 2 more?

Answer 3

Possibly

Answer 4

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.

Answer 5

\[x=\frac{-4\pm\sqrt{-80}}{4}\]?

Answer 6

yes

Answer 7

ok we look for the largest perfect square that is a factor of 80
i get 16 because \(80=16\times 5\)

Answer 8

okay that makes sense

Answer 9

therefore \[\sqrt{-80}=\sqrt{16\times 5\times (-1)}=4\sqrt{5}i\]

Answer 10

woah you lost me D: lol

Answer 11

i guess i skipped a step
\[\sqrt{16\times 5\times (-1)}=\sqrt{16}\sqrt{5}\sqrt{-1}=4\sqrt{5}i\]

Answer 12

since \(\sqrt{16}=4\) and \(\sqrt{-1}\) gets written as \(i\)

Answer 13

that is why we were looking for the largest perfect square that is a factor of 80

Answer 14

let me know when we can continue, it is not over yet

Answer 15

okay that makes sense

Answer 16

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

Answer 17

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 \[\sqrt{-50}=5\sqrt{2}i\]

Answer 18

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?