Question

Given :
\(x^2-x-1=0\)
then : \(x^8 + \frac{1}{x^8}=?\)

Answer 1

have u tried it yet?

Answer 2

For better visual :
\[\large{x^2-x-1=0}\]
\[\large{x^8+\frac{1}{x^8}=?}\]

Yes mukushla.... I tried


1) put x = \(\frac{1}{x-1}\) but that is too long...

Answer 3

@mukushla @hartnn @sauravshakya @ParthKohli

Answer 4

Cant we find the value of x and substitute it

Answer 5

mentioned above but too long...

Answer 6

Well let me know who can solve this

I know the correct soln .... but I will like to know who knows it

Answer 7

Type the answer here

Answer 8

\[x^2-1=x\]\[x-\frac{1}{x}=1\]square both sides 4 time

Answer 9

@allstudentshere : put your answer here

Answer 10

\[x^2 = x +1\]\[x = \sqrt{x + 1}\]
\[x^8 = (x + 1)^4\]\[{1 \over x^8} = {1 \over (x + 1)^4}\]Hmm...

Answer 11

:) only answer friends

Answer 12

1st guess made by mukushla is "wrong"

sorry

Answer 13

lol 47

Answer 14

Yep 47

Answer 15

But that method was not too long

Answer 16

Solve x^2-x-1=0 u will get

Answer 17

And substitute it in x^8+1/x^8

Answer 18

\[\large{x^8 + \frac{1}{x^8}}\]
\[\large{x^8+\frac{1}{x^8}+2-2}\]
\[\large{(x^4+\frac{1}{x^4})^2-2}\]
\[\large{[(x^2+\frac{1}{x^2})^2-2]^2}\]
like this

Answer 19

Guys: mukushla and saurav have right answers

Answer 20

Oh........ But @mukushla method is much better

Answer 21

i did the same thing, still i got -1....maybe i took, -2 instead of +2 somewhere...

Answer 22

\[\large{[[(x+\frac{1}{x})^2-2]^2-2]^2-2}\]
\[\large{\frac{1}{x}=-(\frac{\sqrt{5}+1}{2})}\]

Answer 23

did this...

Answer 24

Again I ask, where did you get +2 and -2 from?

Answer 25

\[\large{x=\frac{-2}{\sqrt{5}+1}}\]
\[\large{\textbf{Put this value in the formula we willl get 47 as answer}}\]

Answer 26

No, I don't mean that. Why did you add and subtract 2 and -2 respectively?

Answer 27

Oh k :
\[\large{(x+\frac{1}{x})^2 = (x^2+\frac{1}{x^2}+2)}\]

hence to make that as look like this... we did subtract 2 and -2 ..

Answer 28

Which formula did you use to complete the square?

Answer 29

x^2-x-1=0
x(x-1-1/x)=0
x-1/x-1=0
x-1/x=1
x^2-2*x*1/x +1/x^2 =1
x^2+1/x^2=3
x^4+1/x^4+2=9
x^4+1/x^4=7
x^8+1/x^8+2=49
x^8+1/x^8=47
@mukushla mentioned this method

Answer 30

\[x^2-x-1=0\]
\[x^{16}-kx^8+1=0\]
\[x^8=y\]
\[y^2-ky+1=0\]
\[y=\frac{k \pm \sqrt{k^2-4}}{2}=x^8\]

Answer 31

\[\frac{k \pm \sqrt{k^2-4}}{2}=(\frac{1 \pm \sqrt{1+4}}{2})^8\]

Answer 32

1+4 = 5

Answer 33

@mathslover the answer is 47 right?

Answer 34

Correct saurav sir :)

Answer 35

And u got it?

Answer 36

Yep.... I knew the soln (and answer) but just wanted to know another way

Answer 37

ahh it's so beautiful to see so many lovers of math/s

Answer 38

including you :) --> lgba --> maths lover

Answer 39

..............

Answer 40

don't ever say that. it disgusts me.

Answer 41

I will like to say again though I care about your feelings....

sorry lgba :'()