Question

If 3+2i is a solution for x^2+mx+n=0, where m and n are real numbers, what is the value of m?

Answer 1

Well, if our coefficients are real and we have a complex root, then the conjugate of the root will be the other root.

Answer 2

Then, we can just set up an equation that will give us 0 when we plug in the roots.
\[ (x - r_1)(x - r_2) = 0 \\
\quad r\mathbf{_1} \text{ is the first root, } \mathbf{r_2} \text{ is the second root.} \]

Answer 3

We then just plug in the roots and expand it out. The imaginary parts will inevitably cancel and we will get the equation in the same form as \(x^2 + mx + n = 0\). We just have to compare the linear coefficient in our expanded equation to the "m" and the constant to "n" for the values of m and n.

Answer 4

?????

Answer 5

Hmm... is there something specific that you don't understand that I can go over clearly?

Answer 6

I dont clearly understand what to do. can you explain just a little simpler please?

Answer 7

Okay... well, do you know the complex root theorem? That, if we have one complex root (a + bi), then the other root is the conjugate / (a - bi)?

Answer 8

Not until now.

Answer 9

Hey, I think just plug X = (3 + 2i ) in!

Answer 10

@Chlorophyll that would be much easier if I had a calculator.

Answer 11

( 3 + 2i)² + n = ( 3 + 2i) m
-> m = 3 + 2i + n/ ( 3 + 2i)

Answer 12

Nope, there's no need calculator at all!

Answer 13

But the answer choices are regular numbers.

Answer 14

Did you check your post for the value of m and n?

Answer 15

what do you mean?

Answer 16

m = - (3 + 2i + n/ ( 3 + 2i) )
If the choice is regular number so they must have value of n.

Answer 17

Ahhhhhhh ! Im so confused.

Answer 18

Alternatively, we could just think about some intuition. If r1 is one solution, and r2 is the other, then this would be true:
(x - r1)(x - r2) = 0 (plug in r1 or r2 and you get 0)
x^2 - r1 x - r2 x + r1 r2 = 0
x^2 - (r1 + r2)x + r1 r2 = 0

comparing this with the original equation "x^2 + mx + n = 0", m = r1 + r2 and n = r1 r2.


We know that r1 = 3 + 2i (given by the problem).
We want an r2 that will both multiply with r1 and add together with r2 to get a real number.
(3 + 2i) + r is a real number
(3 + 2i)r is a real number

Answer 19

m = -(r1 + r2) ***

-(3 + 2i + r2) ***
Should be negative since we have subtraction there. Sorry.

Answer 20

Okay, I think I got that. So far

Answer 21

UGH, my internet is just not helping me at all today. >.<

SO, to get a real number in these cases:
(3 + 2i) + r
(3 + 2i)r
We have to eliminate that "2i". The first case doesn't help us much, we just need a -2i to eliminate the 2i, but we have to look at eliminating in both cases..
For the second case, we need to multiply something to it to completely get rid of the 2i. Well, the only way to do that is with the conjugate, 3 - 2i.

This allows us to deduce that the other root is r2 = 3 - 2i, the conjugate.

Answer 22

Yea, my internet isnt the best either.

Answer 23

So, to conclude, we said that m = -(r1 + r2)
We just plug in the original r1 and the conjugate 3 - 2i and the answer should come out:
m = -((3 - 2i) + (3 + 2i))
= -6

Answer 24

Ughhh.. thank you !!!

Answer 25

that took forever.

Answer 26

Normally, it doesn't take that long to do these sorts of problems. I just kept losing connection. >:[

You're welcome, though. :)

Answer 27

yea, i had to turn my whole computer off and on a couple of times. lol

Answer 28

You can equal to equations:
( 3 + 2i)² + ( 3 + 2i) m + n = ( 3 - 2i)² + ( 3 - 2i) m + n
( 3 + 2i)² + ( 3 + 2i) m = ( 3 - 2i)² + ( 3 - 2i) m
( 3 + 2i)² - ( 3 - 2i)² = [ ( 3 - 2i) - ( 3 + 2i) ] m
==> m = -6

Answer 29

Yep, that would work out as well since the n's cancel.
Once you get the conjugate, there are a lot of ways to find the answer. :D

My method originally was just taking the root and its conjugate in (x - r1)(x - r2), expanding it out, and then comparing the coefficient on x in that to m.

Answer 30

I've never seen that before @Chlorophyll !

Answer 31

@AccessDenied I always extract the idea AFTER you've worked it out :")

Answer 32

@Marketaburn2 Did you see the constant 3 and i terms cancel out?

Answer 33

Yea?

Answer 34

24i = - 4i * m
-> m = -24/4 = -6

Answer 35

Ohh. I got it. :)

Answer 36

Some of my favorite identities about the sum and product of roots come from taking apart the equation like when we found "m = -(r1 + r2)". This is actually always true of the roots for "x^2 + mx + n" (the x^2 coefficient is 1). Similarly, the constant term, n = r1 * r2. There are even similar conclusions in the higher degree polynomials.

I love these very nice conclusions from polynomials. :D

Answer 37

@AccessDenied I agree, the classical conventional method always works :)

Answer 38

I'm sorry, I cant relate. Nothing in math is my favorite. :)

Answer 39

@Marketaburn2 Thanks for posting a pretty interesting complex number problem :)

Answer 40

@Chlorophyll , No problem. Im just reviewing for my Final exam tommorow. :)

Answer 41

That's fine. We all have our opinions. :P
I think I'm one of the only people at my school that actually like math. It's not very 'likeable'. lol

Answer 42

Well, we can't expect the whole world adore you !

Answer 43

@AccessDenied The only time I like math is when I actually understand what Im learning. Which is rare.

Answer 44

*us

Answer 45

Keep your fingers crossed, then!

Answer 46

Yep !

Answer 47

Well, good luck on your final exams. You can do it! :D