TrueTears question thread

[TrueTears]

If a polynomial to the power of 5 has all real coefficients, then what can you say about the solutions?

A. 3 real 2 complex solutions.
B. 5 complex solutions.
C. 5 real solutions.
D. 3 complex 2 real solutions.

thanks

[ryley]

As it has all real coefficients, the conjugate root theorem can be used. From the theorem, the conjugate of any roots must also be a root, so it will have to have an even number of complex roots, so I don't think it can be B or D. Now don't quote me on this, but I'm sure I've read that a polynomial of any degree will have some multiple of two LESS real roots, ie, degree eight, it will have 8, 6, 4, 2 or 0 real roots, so a fifth degree polynomial should have 5, 3 or 1 real root. From this, I think both A and C are possible.

EDIT:
I think I read that last bit here, http://fym.la.asu.edu/~tturner/MAT_117_online/Polynomials_and_their_Functions/polynomial_functions.htm

[TrueTears]

Hmm yeah thanks ryley I　was thinking between A and C as well but answer says it's A, but I don't know why...

[ryley]

Off the top of my head, the only reason I can think of it not being C is how the roots need to be evenly spaced in an argand diagram, can you do that for 5 real solutions?

EDIT: nvm, forgot that the solutions don't have to be unique

[kamil9876]

Now don't quote me on this, but I'm sure I've read that a polynomial of any degree will have some multiple of two LESS real roots, ie, degree eight, it will have 8, 6, 4, 2 or 0 real roots, so a fifth degree polynomial should have 5, 3 or 1 real root. From this, I think both A and C are possible.

That is true. let n be the degree. The number of solutions, including multiplicities(ie getting same factor twice(squared)) will be n=R+C where R is the real solution and C is the non-real solution. Because C must be a multiple of two since they come in pairs we can write it as where is any non-negative integer:


as long as , which is what you claim.

Lulz just realised I quoted against your wishes

[kamil9876]

the roots need to be evenly spaced in an argand diagram?

Sorry for quoting again

This is only true in the equation of the form where is some complex(possibly real)number. However if the polynomial had terms of other degree in there then the evenly spaced thing need not be true.

The statement concerning evenly spread out on argand diagram is derived from assuming it's of the form (you can derive this by writing it and notice that:





Notice how the argument increases by each time. This is why the numbers are all seperated by same angle.

The above reasoning cannot be applied to other kinds of polynomials.

[ryley]

Thanks for correcting that kamil, I was just playing around on the CAS and I realised how fucked up what I said was, can't believe (and not sure why) I thought that extended to all polynomials.

[TrueTears]

Hi here's a question which I think the answer is wrong but can someone please confirm? Thanks!




You don't need to know part a) - c) to do d).

So d) clearly says the FORCE is inversely proportional which means where

Now using Newton's second law:

Here's the answers:



Why do did they leave out the mass Michael? Are they wrong?

[TrueTears]

Also another question, how do you do it?




Thankee

[TrueTears]

Last one:

Part e) is what I'm having trouble with, I can do part e) 2 different ways but they give an answer which is 0.2 off each other but I can't see a flaw in either of the 2 methods so can someone please check?



Very simple



d)

Now part e).
First method: Closest to the plane is when the dot product of and [Straight line is the fastest route]



Solve for t using a calc yields t = 5.31 min (2 dp)

Second method: Closest to the plane is when the magnitude of is a minimum, thus whatever is under the square root must be a minimum.

is a minimum when is a minimum.

So, and set to 0 and solve for t yields t = 5.50 min (2 dp)

Now question says to nearest minute, well method one would be 5 min and method 2 would be 6 min. But both ways seem correct to me? Why is there a discrepancy between the 2 methods?

[kamil9876]

Now part e).
First method: Closest to the plane is when the dot product of and = 0 [Straight line is the fastest route]

Not true. When he is at the point closest to A the direction he is facing is perpendicular to AP (you also need to test endpoints of domain because they could be even less(beginning/ end of journey), but I assume you have ruled them out).

Hence you need to find the dot product between AP and his direction vector (velocity vector).

[TrueTears]

Now part e).
First method: Closest to the plane is when the dot product of and = 0 [Straight line is the fastest route]

Not true. When he is at the point closest to A the direction he is facing is perpendicular to AP (you also need to test endpoints of domain because they could be even less(beginning/ end of journey), but I assume you have ruled them out).

Hence you need to find the dot product between AP and his direction vector (velocity vector).

I have no idea what you're talking about.

[kamil9876]

Sorry just realised my method is the same as yours(because the cartesian equation is a straightline). My method is just more general for any curve.

[TrueTears]

Sorry just realised my method is the same as yours(because the cartesian equation is a straightline). My method is just more general for any curve.

Yes that's what I thought... so... what's the wrong thing in the 2 methods...?

[kamil9876]

on second thought. The cartesian equation of OP is not a straight line:



But the above must be constnat for a striaght line going through origin. Hence the thing is not a straight line so method1 doesn't work. THe more general variation of it should though.