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#2 Help Me ! » Very simple equation with logarithms » 2007-09-16 01:57:35

Jitse
Replies: 2

I've got some problems with the logarithms, this is the equation:

I could do something like:

Or maybe:

But I just don't know how to get to this result:


They said something like taking 10 as exponent or so, but I really don't see it.

Thanks in advance

#3 Re: Help Me ! » Quartic and quintic functions » 2007-09-16 00:34:56

Sorry, JaneFairfax was faster, hehehe.

JaneFairfax wrote:

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. neutral

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.)

#5 Re: Help Me ! » Quartic and quintic functions » 2007-09-15 03:42:31

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

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

#6 Re: Help Me ! » Simplification of equation » 2007-09-15 03:25:18

roll Sometimes I can be so blind, lol. Thanks. I'll just have to start exercising algebra some more, so I get used to all the common rules like this one.

#7 Help Me ! » Simplification of equation » 2007-09-15 03:04:40

Jitse
Replies: 2

This is an equation, which I can solve, but my result is quite complex, and I know you could easily simplify it to a better-looking result, I just don't know how.

This is the equation:

Now how I solve this:





And then I take my calculator, and get the result:


But I know the more decent result (equal to the one above) would be:


So my question is, how do you get from

to that good-looking result? Or did I just do it in a weird way?

Thanks in advance

#8 Help Me ! » Quartic and quintic functions » 2007-09-15 02:41:17

Jitse
Replies: 7

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

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