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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.
B.
And the solutions are:
A.
3 and -3
B.
1,
Thanks in advance
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The advice for question A is good.
If you do that, then you get
.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.
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I can get the results -1 and 9 when you do
, but how do you then substitute this into the original one?Offline
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.
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Ok, thank you very much.
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B.
B.
and
1,
That cant be right, can it? A quintic polynomial ought to have an odd number of real roots; you cant have just two real roots here.
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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.
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Sorry, JaneFairfax was faster, hehehe.
That cant be right, can it? A quintic polynomial ought to have an odd number of real roots; you cant 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:
Final results:
Last edited by Jitse (2007-09-16 00:37:53)
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