Quadratic with solutions???

Question

Ok, I have no idea how to solve this problem...

Find a quadratic with solutions: 3 + i, 3 - i, 2

I have no idea.
Can someone please explain this to me?

Answer 1

(x-2)(x-3+i)(x-3-i)
This is your function factored. Multiply it out and you get
x^3 -8x^2 +22x -20.This is not a quadratic function, but a cubic. A quadratic has only 2 roots or solutions.

Answer 2

Since the solutions are x = 3 + i and x = 3 - i and x = 2, just work backward.

x = 3 + i and x = 3 - i and x = 2 so,

x - (3 +i) = 0 and x - (3 - i) = 0 and x - 2 = 0

(x - (3 + i))(x - (3 -i))(x - 2) = 0

now just multiply it all out, collect like terms and you have it.

Answer 3

Go back to how you come to the solutions for the quadradic equations....

If you were to solve one, you basically factor it and come to
x+3=0 or x-4=0 then solve for x correct?

Then, what would you have to do to go backwards? You have the solutions already. Can you put it in something = 0 format?

If you had to factor the formula to get there, what would you do to go backwards?

What's i? How do you get i? (imaginary number) Go back to the definition of the imaginary number.. What does i^2 equal to?

Answer 4

whenever they give you the roots just put it in this format:
say the roots are 1, 2, and -3.

then the equation would be

(x-1)(x-2)(x+3)=0

get it?