# Imaginary numbers entertwined with quadratics

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

-Hoot

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

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

Hootenanny
Staff Emeritus
Gold Member
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

Last edited:
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... im confused...

dav2008
Gold Member
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.

My davey jones's locker i think ive 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
Staff Emeritus
Gold Member
Are you correct?

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

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

VietDao29
Homework Helper
runicle said:
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? :)