Here is a tough question (at least it was tough for me .

Question

Here is a tough question (at least it was tough for me >.<)

anonymous · Answer

attachment

anonymous · Answer

I do have a solution typed out in LaTex, if anyone cares to see it.

anonymous · Answer

elementary symmetric functions?  or some algebra?  curious

anonymous · Answer

Give me a minute...this is a lot more than elementary functions....

anonymous · Answer

My solution has a lot of algebra, but i never really thought of looking at it through elementary symmetric functions, so there might be a solution through that venue.

anonymous · Answer

well i am clueless. hint?

anonymous · Answer

Ok I'll give it a shot...

Four roots, we'll call them a, b, c, d.

The sum of the roots of a polynomial is -b/a, where b is the coefficient of the term one degree less than the first term and a is the coefficient of the leading term. 

We have

$$x ^{4}-18x ^{3}+kx ^{2}+200x-1984=0$$

The sum of the zeroes is -(-18)/1=18

So a+b+c+d=18


Right track so far?

anonymous · Answer

Thats right, and also in the right direction. Looking at the coefficients as a "product" of the roots (not product as in multiplication, but product as in the roots determine the coefficients)

anonymous · Answer

I guess I should have said this in the beginning, this isnt a "fast answer" type of problem, its one you really need to sit down and think about.

anonymous · Answer

Yeah...I kind of figured that after a while :)

anonymous · Answer

I'm trying to figure out the relationship between a, b, c, d and k...

anonymous · Answer

k=86

anonymous · Answer

robtobey is correct o.O awesome job, i wont even say how long it took me to figure it out >.<

anonymous · Answer

Ok...I think I might have it...

If we think about a cubic polynomial, i.e. 

$$at ^{3}+bt ^{2}+ct+d=0$$

with roots x, y, and z...


We have the following properties:

1) x+y+z=-b/a, as I noted above.

2) xy+xz+yz=-c/a

3) xyz=-d/a

Logically, this should be expanded to quartic polynomials.

So for the polynomial 

$$x ^{4}-18x ^{3}+kx ^{2}+200x-1984=0$$

We have

a+b+c+d=18

and from the above properties, we also have:

ab+ac+ad+bc+bd+cd=k

abc+abd+acd+bcd=-200

abcd=-1984

We let ab=-32

anonymous · Answer

Then we substitute t=a+b, u=c+d, and v=cd

This yields:

t+u=18

k=-32+tu+v

-32u+tv=-200

-32v=-1984

So v=62

Which also gets us t=4, u=14

so we have

k=-32+(4)(14)+62=62+56-32=86

I'm getting 86 as well...

anonymous · Answer

Yes, thats right, i'll post my solution as well.  Its extremely formal so it might seem long, sry bout that, it was for a Problem Solving class where we had to type formal proofs.

anonymous · Answer

Solved by trial and error with the aid of Mathematica Home Edition.  Refer to the attachment.  Each solution takes less than 3 milliseconds on a mid 2010 IMac.