## livelaughlilz 3 years ago find a quadratic equation whose roots are 3+i and -1/2

1. precal

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

2. precal

this is really a cubic

3. vipul92

right this que. is wrong

4. precal

were you asked to find a polynomial?

5. precal

if so then, we can do it

6. livelaughlilz

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?

7. CliffSedge

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

8. CliffSedge

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

9. livelaughlilz

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

10. livelaughlilz

@cliffsedge hello...?

11. CliffSedge

(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 $\large x^2-\frac{5}{2}x-ix-\frac{3}{2}-\frac{1}{2}i$

12. livelaughlilz

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

13. CliffSedge

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.

14. livelaughlilz

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

15. CliffSedge

I don't know. That looks pretty weird; I'm not even sure it's quadratic.

16. precal

not quadratic, let's review the 3 possible solutions of a quadratic|dw:1350688342898:dw|one solution double root

17. precal

|dw:1350688368004:dw|2solutions

18. precal

|dw:1350688390750:dw|no solution 2 complex solutions not a quadratic, no matter how you look at it

19. CliffSedge

@precal , what if the quadratic equation has complex coefficients like the ones we found above?

20. precal

not possible, complex conjugate theorem states that complex solutions come in pairs. Even if you are not given both, you can assume the other by the thm if 3+i is a known zero, we can assume 3-i because of the thm.....this is not a quadratic (This would be a great counterexample of a quadratic)

21. CliffSedge

The complex conjugate theorem only applies to polynomials with real coefficients @precal

22. precal

23. CliffSedge

That's right. The complex conjugate theorem as it applies to quadratic polynomials only applies to quadratic polynomials with real coefficients.

24. precal
25. precal

I think you are misinterpreting the thm..........either way it is not a quadratic I suspect it is a cubic then it would make sense

26. CliffSedge

Nope. The theorem states that *if* a polynomial has real coefficients then complex roots will come in conjugate pairs. If the polynomials does not have real coefficients then the theorem does not apply.

27. CliffSedge

$\large x^2-\frac{5+2i}{2}x-\frac{3+i}{2}=0$ Is a quadratic equation, and it does have roots {3+i,-1/2} QED.

28. precal

QED that is only used in formal proofs, trust me you did not provide a formal proof.....still not a quadratic

29. CliffSedge

I was just being cute with the "QED" And yes it is a quadratic equation. Can you provide any sources to back up your claim that it is not @precal ?

30. precal
31. precal
32. precal

I could go on but then that would be me being "cute" bottom line is complex solutions come in pairs..............If you know one, you can assume the other.....