## anonymous 4 years ago 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.

1. AccessDenied

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

2. AccessDenied

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. :)

3. anonymous

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

4. AccessDenied

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

5. anonymous

does "i" always equal -1 ?

6. AccessDenied

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

7. anonymous

ohh i get it! thanks!

8. AccessDenied

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

9. anonymous

im going to try it now

10. anonymous

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

11. AccessDenied

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?

12. anonymous

yes!!

13. anonymous

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?

14. AccessDenied

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

15. anonymous

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

16. AccessDenied

You're welcome! :)