Question

if x+x^-1=5, then find the value of x^4 + x^-4

Answer 1

find x then substitute

Answer 2

no need for that, try expanding this and see what you get:\[\left(x+\frac{1}{x}\right)^2=\]

Answer 3

is it possible to solve in an another way?

Answer 4

\[x ^{2}+(1/x ^{2})\]

Answer 5

not quiet, remember that:\[(a+b)^2=a^2+2ab+b^2\]

Answer 6

oh there will be an extra 2

Answer 7

yes, so you get:\[\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\]therefore:\[5^2=x^2+\frac{1}{x^2}+2\]

Answer 8

now simplify to get:\[x^2+\frac{1}{x^2}=..\]then apply same rule again and you're done

Answer 9

ok let me try

Answer 10

ok done. thankyou very much

Answer 11

yw :)

Answer 12

can we simply solve this equation ?

x+1/x=5

x^2-5x+1=0

http://prntscr.com/4gafv8

@rational

Answer 13

even though i liked the @asnaseer answer

Answer 14

you can, but squaring twice /seems/ simpler - I guess its a matter of taste? :)

Answer 15

ik :P
it was just a though, cuz we might have 2 answers ( i dint check yet :P)

Answer 16

solving x is a good idea too as x and 1/x magically give you conjugate pairs here

Answer 17

x^4 + 1/x^4 = 5^4 - 2

Answer 18

ohh both give same solution , and one solution cool :D

Answer 19

hmm n0
its 23^2-2

Answer 20

how did u work it ?

Answer 21

if you try and solve the quadratic you get:\[x=\frac{5\pm\sqrt{21}}{2}\]both of these lead to:\[x^4+\frac{1}{x^4}=527\]
If you use the method I described, then you first get:\[x^2+\frac{1}{x^2}=23\]and then:\[\left(x^2+\frac{1}{x^2}\right)^2=23^2\]therefore:\[x^4+\frac{1}{x^4}+2=23^2\]therefore:\[x^4+\frac{1}{x^4}=23^2-2=527\]

Answer 22

exactly! solving x explicitly requires binomial expansion i guess
http://www.wolframalpha.com/input/?i=1%2F2%5E4%282%285%5E4%2B%5Cbinom%7B4%7D%7B2%7D*5%5E2*21%2B21%5E2%29++%29

not sure if there is a way to avoid expansion if we solve x explicit

Answer 23

solving x is quadrati equation :o