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Jitse

Member

Registered: 2007-09-15

Posts: 6

Quartic and quintic functions

I know how to solve linear, quadratic and cubic functions, but I get stuck at the quartic and quintic functions, hehe. Could someone please help me out? I've got a few problems here, with the solutions, but I don't know how solve them. These are just secundary school questions though.

A.

They were saying something about take

as y and then just do what you'd normally do.

B.

Here they said you should use the Horner thing, and then you'd have a quartic function, with which you should just do the same as with A.

And the solutions are:

A.
3 and -3

B.
1,

and

Thanks in advance

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quartic and quintic functions

The advice for question A is good.

If you do that, then you get

.
Solving for y gets you 9 and -1.

You then substitute those back into y=x². For y=9, x=±3 and for y=-1 there are no real values of x. So 3 and -3 are your two answers.

I don't know what the Horner thing is, so I can't help you for B.

---

Why did the vector cross the road?
It wanted to be normal.

Jitse

Member

Registered: 2007-09-15

Posts: 6

Re: Quartic and quintic functions

I can get the results -1 and 9 when you do

, but how do you then substitute this into the original one?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quartic and quintic functions

Just replace y with those answers.

So you have 9 = x², which gives solutions of 3 and -3.
Then you have -1 = x², which has no real solutions.

---

Why did the vector cross the road?
It wanted to be normal.

Jitse

Member

Registered: 2007-09-15

Posts: 6

Re: Quartic and quintic functions

Ok, thank you very much.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Quartic and quintic functions

B.

B.
1,

and

That can’t be right, can it? A quintic polynomial ought to have an odd number of real roots; you can’t have just two real roots here.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Quartic and quintic functions

I see what you mean. Find an obvious real root (in this case x = 1), then divide it out using the Horner method.

And now you solve the quartic.

Jitse

Member

Registered: 2007-09-15

Posts: 6

Re: Quartic and quintic functions

Sorry, JaneFairfax was faster, hehehe.

That can’t be right, can it? A quintic polynomial ought to have an odd number of real roots; you can’t have just two real roots here.

Yes I'm sorry, I forgot to mention the third result: 1.

I now know how to solve that quintic function by the way. The Horner scheme is just a faster way to divide a polynomial. Like this:

Now you write all coefficients in a scheme:

   |   1    -1    -1    1    -6    6
   |
   |----------------------------------

Then you take one number that gives 0 as result when you replace the x of the function with it, so for example when you take 1:

You add that number to the scheme:

   |   1    -1    -1    1    -6    6
1  |
   |----------------------------------

And now you do the specific algorithm (Check this: http://en.wikipedia.org/wiki/Horner_scheme).

   |   1    -1    -1    1    -6    6
1  |         1     0   -1     0   -6
   |----------------------------------
   |   1     0    -1    0    -6  | 0

The new numbers are the coefficients of the new function, so:

Then you just do the same as you did with question A.

Now find the determinant:

Use the formula:

On y1:

On y2:

Substitute:

-> No real solutions.

Final results:

-> (The one of the Horner scheme.)

Last edited by Jitse (2007-09-16 00:37:53)