Imaginary numbers entertwined with quadratics

[runicle]

Hi, I'm not sure if this is calculas based or algebra based so here's the question.

For each pair of roots find a quadratic equation. Express your equation in the form of ax^2+bx+c=0

(
(A) 2i, -2i

For this question i don't know what is being asked so i guess the pairs could be x...
I know this:
ax^2+bx+c, a(x-h)^2+k and a(x-s)(x-t)
So the problem is how can i use the things that i know to answer the question. If you can please use analogies and don't give me the answer i just need to know what to do.

 

[Hootenanny]

Think about the process you go through to solve a quadratic and apply it in reverse. Take note of the complex numbers though.

-Hoot

Note to mentors/admin suggest moving this thread to pre calc forum.

 

[runicle]

Im trying this... I just used (a+bi)(c+di)
-(a+bi)(c+di)=0
-ac+adi+bci+bdi^2=0
-from previous question 2i, -2i I represented a = 0... and so forth
-

 

[runicle]

Although i don't know what 2i represents... Is it x b c?

 

[Hootenanny]

From what I can see the solution is pretty simple

$$(x - (2i))(x - (-2i)) = 0$$

or am I missing something?

Remember that $i^2 = -1$

[EDIT] I've edited my post to change my stupid mistake

 

[runicle]

I doubt that hootenay... I'm thinking of x = -b -+ Square root of b^2-4ac over 2a equation and deriving it so that something will occur... I am confused...

 

[dav2008]

Like hoot implied(I think slightly incorrectly), it's always true that if r1 and r2 are the two roots to a quadratic then the characteristic quadratic equation is (x-r1)(x-r2)=0. Then you just expand that to get it into the standard form.

The same argument can be extended to higher degree polynomial equations.

 

[runicle]

My davey jones's locker i think I've got It!
Correct me if I'm right
0=(x-(2x))(x-(-2x))
0=(x-2i)(x+2i)
0=(x^2+2ix-2ix-4i)
0=(x^2+4)
am i correct?
now if the pairs were 1+3i, 1-3i I would have sub the same way like
0=(x-(1+3i))(x-(1-3i))
0=(x-1+3i)(x+1-3i)
and solve from that correct?

 

[Integral]

Are you correct?

It is very easy to check your work. What are the roots of

$$x^2 + 4 =0$$ ?

 

[VietDao29]

now if the pairs were 1+3i, 1-3i I would have sub the same way like
0=(x-(1+3i))(x-(1-3i))
0=(x-1+3i)(x+1-3i)
and solve from that correct?

You made a small mistake when going from the first line to the second line. Your second line should reads:
0 = (x - 1 - 3i) (x - 1 + 3i).
Now just expand all the terms out, simplify it, and it's all done.
Can you go from here? :)