If a + bi is a root of the quadratic equation x^2 + cx + d = 0, then show that a^2 + b^2 = d and 2a+c=0.

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If a + bi is a root of the quadratic equation x^2 + cx + d = 0, then show that a^2 + b^2 = d and 2a+c=0.

Mathematics
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I believe that we could use quadratic formula here, and equate the like-parts (real and imaginary): \( \Large { x = \frac{\neg b + \sqrt{b^2 - 4ac}}{2a} } \) We'll only need the positive part because x = a + bi, it is adding the bi. So, this equation would look like this: \( \Large a + bi = \frac{\neg c + \sqrt{c^2 - 4d}}{2} \) To get the imaginary part, we need to factor out a -1 from \(c^2 - 4d\): \( \Large a + bi = \frac{\neg c + i \sqrt{4d - c^2}}{2} \) \( \Large a + bi = \neg \frac{c}{2} + \frac{\sqrt{4d - c^2}}{2} i \)
So, if we equate the like-parts here, we get: a = -c/2 b = sqrt(4d - c^2)/2 (The i's can divide off) From there, it takes some clever algebra work to get the two results you wanted to show. :)
okay that makes sense, but can you explain the "i", i dont understand why - 1 gets the imaginary part??

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we factor out the -1 from under a square root, so the square root of the -1 = i \( \sqrt{c^2 - 4d} = \sqrt{(-1)(4d - c^2)} = \sqrt{-1}\sqrt{4d - c^2}\)
does "i" always equal -1 ?
\(i =\sqrt{-1}\), not just -1.
ohh i get it! thanks!
You're welcome! :) Were you able to figure out the rest?
im going to try it now
for quadratic formula, isnt its x= -b +/- square roots... why did you just write + ?
We're only dealing with one root when we use 'x = a + bi,' the positive square root here. The other root is 'x = a - bi', which is where the negative square root is used. Does that make sense?
yes!!
i got a^2 + b^2 = d, but i cant get 2a+c =0 2(-c/2) +\[\sqrt{4d-c ^{2}}\] =0 -c + d-c = 0?
hm... a = -c/2 Here we should have enough to get 2a + c = 0, we have all our variables in an equation! 2a = -c we can add c to both sides 2a + c = 0
yeaah! I didnt see that, I got it thanks so much!
You're welcome! :)

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