find a quadratic equation whose roots are 3+i and -1/2

Question

precal · Answer

can't do this because complex come in pairs
3+i and 3-i are solutions

precal · Answer

this is really a cubic

anonymous · Answer

right this que. is wrong

precal · Answer

were you asked to find a polynomial?

precal · Answer

if so then, we can do it

anonymous · Answer

no i was asked to find a quadratic. wouldn't the 3 + i and 3 - i pair only be required if i was asked for integral coefficients?

anonymous · Answer

If there is no requirement to have rational coefficients, then there's no problem, just multiply the factors together.

anonymous · Answer

*(meant to say "no requirement to have 'real' coefficients, ...")

anonymous · Answer

i know the formula x^2 - (sum of the roots)x + product of the roots = 0. but i did it and got x^2 - (5/2 + i)x - 2i = 0. the answer is supposed to be x^2-5+2i/2x-3+i/2

anonymous · Answer

@cliffsedge hello...?

anonymous · Answer

(Sorry, didn't see your reply)
If the roots are 3+i and -1/2, the the factors are (x-3-i) and (x+1/2)
When I multiply those together, I get 
$$\large x^2-\frac{5}{2}x-ix-\frac{3}{2}-\frac{1}{2}i$$

anonymous · Answer

I got that too. But the answer is supposed to be x^2-5+2i/2x-3+i/2

anonymous · Answer

I'm not sure I understand that answer. You mean this?
$$\large x^2-\frac{(5+2i)}{2}x-\frac{(3+i)}{2}$$

If so, then they are equivalent, try simplifying the above.

anonymous · Answer

no i meant this: |dw:1350666676396:dw|