Eyes on Final Fantasy Forums > Reference > Study Hall > Maths help...

[Tavrobel]

Nuke's method is fine, and as far as I can see, the only way of getting to the answer (unless you know what I mean by minding your Ps and Qs). Just clean up your work a little bit and you should be fine.

If this is going up on the board, then all of the best of luck to you. Alternatively, you can Power Rule the equation a couple of times, and eventually, you will get down to zero. It's fun. I do this all the time to freak out the underclassmen.

This was my original work until I did it the other way:
Multiply both sides by x² (similar to Mulley's thing, without fractions).
(x² + x + x^-1 + x^-2 = 4)x².
x⁴ + x³ + x + 1 = 4x²

Bring 4x² to the other side, and set it equal to zero.
x⁴ + x³ + (-4x²) + x + 1 = 0
Factor out x³ from the first two terms (you can do this by the way).
x³(x + 1) + (-4x²) + x + 1 = 0

Stick a parentheses around common terms, and group together:
x³(x + 1) + (x + 1) + (-4x²) = 0
Factor out (x + 1):
(x + 1)(x³ + 1) - 4x² = 0
Solve for zero. This part pretty much needs Ps and Qs.

[I Took the Red Pill]

It's pretty straightforward if you use Newton's Method, you'd have to convince your teacher that you recognize the decimal as (-3 ± √5)/2 though

[Tavrobel]

Only if she knows her square roots of numbers up to ten to three decimals. (2.236)

They could also be equivalent fractions. I doubt she knows Calculus at all.

[Flying Mullet]

That's what I missed. Once I had ((x³ + 1)(x + 1)) / x² = 4, I needed to multiply both sides by x² and subtract 4x² from both sides. Then I could solve for 0.

*smacks forehead*

[Selena_Akariko]

Um... Thanks for your help, people, but... Well, first of all, I have my maths in Russian, and those English math terms are a bit confusing. And secondly,

I have asked my maths teacher today, and she solved that problem in less than a minute.

I'm sure the simplicity of it will surprise you all.

Here's how you actually supposed to do that:

x² + x + x^-1 + x^-2 = 4

(x² + 1/x²) + (x + 1/x) = 4

It's common knowledge that (x + 1/x)² = x² + 2 + 1/x²

Therefore x² + 1/x² = (x + 1/x)² - 2

(x + 1/x)² - 2 + (x + 1/x) = 4

(x + 1/x)² + (x + 1/x) - 6 = 0

Then we say that (x + 1/x) = t and solve the t² + t - 6 = 0 quadratic equation.

Sorry for bothering all you people ^^ But this way of solving might be useful if you ever come across something like that again.

Thank you all again for bothering to help me

Selena

[blackmage_nuke]

That method seems abit too drawn out, you might save a few lines of working but in the end when you have a thousand of these little methods to try remember by the time you can think of an appropriate one you couldve probably finished it using the long way (not to mention the time it may require to check if the method will work so you dont get a dead end). So it doesnt seem much easier in exam conditions unless your good at that sort of thing (which i guess you should be)