Polynomials (1 Viewer)

[mojako]

Would somebody please point out the errors in the following attachment?
Thanks

 

[:: ck ::]

q1 - ur expansion when u "squared both sides" [near the end] is wrong

that said, im not sure if the answer is wrong becoz ull end up with x^6 in ur ans even if u did expand it properly

unless i missed an earlier errror in ur working

using the correct expansion (hopefully) i get :

x⁶ -2x⁵ -47x⁴ - 48x³ + 96x² -256x + 256

mayb sum1 can confirm this...

q2 - use the identity sin²@ = 1-cos²@

i.e : cos4@ = cos⁴@ -6cos²@(1-cos²@) + (1-cos²@)²

 

[mojako]

using the correct expansion (hopefully) i get :

x⁶ -2x⁵ -47x⁴ - 48x³ + 96x² -256x + 256

Isn't that what I wrote???
I actually was re-doing the expansion while you edited your message... and still got the same result.

I *think* if it's done properly we should get the square of the cubic polynomial, which makes the roots have multiplicity of 2 (or more if the original cubic already has double or triple root).

 

[:: ck ::]

hrmm sorry i didn read ur thing properly... i looked at ur "squaring both sides" part and used a calcuilator to sub a value for x and see if both sides equaled the square of the line above it.. and it wasnt the same...

so i assumed ur last answer was wrong and wrote down wot i got on paper without looking at ur answer

hrmm ill do the question later on in the day and post it up

 

[mojako]

Thanks for giving a sacrifice of your valuable time, hehehe

 

[CM_Tutor]

For Q1, you have x³ + x² + 2 = 0, with roots alpha, beta and gamma. You want an equation with roots alpha⁴, beta⁴ and gamma⁴. So, let u = x⁴, and so x = u ^1/4. Substituting, we get:

u^3/4 + u^1/2 + 2 = 0
u^3/4 = -(2 + u^1/2)
On squaring both suides, this becomes u * sqrt(u) = 4 + 4 * sqrt(u) + u
(u - 4) * sqrt(u) = u + 4
Squaring both sides again, we get u(u² - 8u + 16) = u² + 8u + 16

Hence, the equation u³ - 9u² + 8u - 16 = 0 has the required roots

 

[Wohzazz]

Sorry, i was wondering how you managed to write mathematical symbols in a document like theta.

 

[J0n]

Originally posted by Wohzazz
Sorry, i was wondering how you managed to write mathematical symbols in a document like theta.

In Microsoft Word, Insert->Object->Microsoft Equation

 

[mojako]

Originally posted by Wohzazz
Sorry, i was wondering how you managed to write mathematical symbols in a document like theta.

In addition to J0n reply, may I add:
You need to have installed the Microsoft Equation Editor for that option to appear.

Go to Control Panel then Add or Remove Programs then look for Microsoft Office or Microsoft Word then click on "Change" if you have WinXP or "Uninstall" (I think) if Win98. It won't uninstall the program as soona s you click, though. In the upcoming window, make sure Add or Remove features is selected, then click Next then give a check mark on "Choose advanced customization of applications" or similar kind of words.
Now, click on the + sign next to "Office Tools" and click on Equation Editor and choose Run from My Computer.

Alternatively, I *think* you can open any document containing an equation object (eg my attachments here). Right click on the equation > Equation Object > Edit. And Word should automatically install Equation Editor.

Another alternative if you don't want to install Equation Editor OR you've lost your Office CD somehow (and I'm not sure if you can download the Equation Editor from the internet), you can write a lot of equations as a "Field". Some mathematical symbols are available in Insert > Symbols. But this Field business is not very convenient to deal with.

 

[mojako]

Originally posted by CM_Tutor
For Q1, you have x³ + x² + 2 = 0, with roots alpha, beta and gamma. You want an equation with roots alpha⁴, beta⁴ and gamma⁴. So, let u = x⁴, and so x = u ^1/4. Substituting, we get:

u^3/4 + u^1/2 + 2 = 0
u^3/4 = -(2 + u^1/2)
On squaring both suides, this becomes u * sqrt(u) = 4 + 4 * sqrt(u) + u
(u - 4) * sqrt(u) = u + 4
Squaring both sides again, we get u(u² - 8u + 16) = u² + 8u + 16

Hence, the equation u³ - 9u² + 8u - 16 = 0 has the required roots

Thanks again for your answer...
But what's wrong with the steps I did (in the Word attachment)?

 

[CM_Tutor]

You have produced a degree 6 polynomial with six roots alpha⁴, -alpha⁴, beta⁴, -beta⁴, gamma⁴, -gamma⁴. This has occured because you raised to the fourth power in going from line 2 to line 3 of the solution in ask.doc. You should have simply squared.

 

[mojako]

okay...
thanks.