Maths help...

[Selena_Akariko]

x<SUP>2</SUP> + x + x<SUP>-1</SUP> + x<SUP>-2</SUP> = 4

I have been trying to solve this for more than 3 hours now, but I have no idea how to do it I've tried lots of different ways, but none of them worked.

So... Help? Please?

Selena
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[Kirobaito]

x = 1

lol

As for how to solve it, I have no idea... but the answer is x = 1.

[Flying Mullet]

That's a doozy. I rewrote the problem as:
x<SUP>2</SUP> + x + 1/x + 1/x<SUP>2</SUP> = 4
then converted everything to a common denominator of x<SUP>2</SUP> which gave me this:
(x<SUP>4</SUP> + x<SUP>3</SUP> + x + 1) / x<SUP>2</SUP> = 4
but that's as far as I got.

I'm sure there's some identity or such that can be applied here, I just can't think of any that fit the bill.

EDIT:
You can then factor the numerator into:
(x<SUP>3</SUP> + 1)(x + 1)
which you can cut into:
(x<SUP>3</SUP> + 1)/x * (x + 1)/x = 4

[rubah]

i'd graph it imo

-2.618024
-.381966
+1.00 (.99999999965 yeah I dunno about that either!)

Note that there is a hole in the graph at x=0 since you have to divide by x, lol not happenin'!

also there's a good chance I might have graphed it wrong, and a definite chance that there's some fractional representation of those decimals

[Selena_Akariko]

Kirobaito: Wish it was that easy... Our teacher requires the solving, not the result

Flying Mullet,

(x<SUP>3</SUP> + 1)(x + 1)
which you can cut into:
(x<SUP>3</SUP> + 1)/x * (x + 1)/x = 4

Well, I tried that, but it doesn't seem to give any hints about how to solve it...

This problem is supposed to turn out very easy once you find the way how to do it... At least, that's how other problems in that textbook worked.

rubah, I'm afraid graphing isn't an option

Selena

Edit: The answer is supposed to be: x1 = x2 = 1; x3,4 =(-3 ± √5)/2

[rubah]

well, the thing about graphing is that once you have your answers, you can then start to work backwards sometimes, and then you'll realize how to do it. that's how I did a lot of trig homework xD

Anyways, -2.61etc *2 +3 +sqrt5 happens to be 0 so I'm pleased that that answer was correct
-.381etc * 2 +3 -sqrt5 is also zero 8)

sorry i can't help you do it algebraicly

[Flying Mullet]

Kirobaito: Wish it was that easy... Our teacher requires the solving, not the result

Flying Mullet,

(x<SUP>3</SUP> + 1)(x + 1)
which you can cut into:
(x<SUP>3</SUP> + 1)/x * (x + 1)/x = 4

Well, I tried that, but it doesn't seem to give any hints about how to solve it...

This problem is supposed to turn out very easy once you find the way how to do it... At least, that's how other problems in that textbook worked.

rubah, I'm afraid graphing isn't an option

Selena

Edit: The answer is supposed to be: x1 = x2 = 1; x3,4 =(-3 ± √5)/2

I didn't figure out how to solve it. I was hoping that it might help out in some way.

[blackmage_nuke]

This might sound like a noob method but i moved the 4 to the left side to get
x^2+x+1/x+1/x^2-4=0
then i did the same thing as flying mullet and made one denominator
(x4 + x3 + x + 1 - 4)/x2 = 0

So i just ignored that bottom x2 because i can move it to the other side
Then i tested values for x4+x3+x+1-4 until it = 0 to find a factor
1 was a factor
(x-1)(x3+2x2-2x-1)=0
Then i subed for x3+2x2-2x-1 until i found a factor
1 was a factor
(x-1)(x-1)(x2+3x+1)
from here u get x=1 or x2+3x+1=0 which u can use the quadratic formula for x = (-3+root5)/2

Ive never been good at remembering or applying any of the larger theorems or rule so this might not be the quickest way

[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:

x<SUP>2</SUP> + x + x<SUP>-1</SUP> + x<SUP>-2</SUP> = 4

(x<SUP>2</SUP> + 1/x<SUP>2</SUP>) + (x + 1/x) = 4

It's common knowledge that (x + 1/x)<SUP>2 </SUP>= x<SUP>2</SUP> + 2 + 1/x<SUP>2 </SUP>
<SUP></SUP>
Therefore x<SUP>2</SUP> + 1/x<SUP>2</SUP> = (x + 1/x)<SUP>2</SUP> - 2

(x + 1/x)<SUP>2</SUP> - 2 + (x + 1/x) = 4

(x + 1/x)<SUP>2 </SUP>+ (x + 1/x) - 6 = 0

Then we say that (x + 1/x) = t and solve the t<SUP>2 </SUP>+ 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)