Question

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.

Answer 1

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 \(c^2 - 4d\):
\( \Large a + bi = \frac{\neg c + i \sqrt{4d - c^2}}{2} \)
\( \Large a + bi = \neg \frac{c}{2} + \frac{\sqrt{4d - c^2}}{2} i \)

Answer 2

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. :)

Answer 3

okay that makes sense, but can you explain the "i", i dont understand why - 1 gets the imaginary part??

Answer 4

we factor out the -1 from under a square root, so the square root of the -1 = i
\( \sqrt{c^2 - 4d} = \sqrt{(-1)(4d - c^2)} = \sqrt{-1}\sqrt{4d - c^2}\)

Answer 5

does "i" always equal -1 ?

Answer 6

\(i =\sqrt{-1}\), not just -1.

Answer 7

ohh i get it! thanks!

Answer 8

You're welcome! :)
Were you able to figure out the rest?

Answer 9

im going to try it now

Answer 10

for quadratic formula, isnt its x= -b +/- square roots...
why did you just write + ?

Answer 11

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?

Answer 12

yes!!

Answer 13

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?

Answer 14

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

Answer 15

yeaah! I didnt see that, I got it thanks so much!

Answer 16

You're welcome! :)