Given #x=a+1/a# and #y=a-1/a# calculate #5(x^2+y^2)^2# ?

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Answer 1 · 2017-03-10T11:28:35

given
#x=5^(1/4) + 5^(-1/4)#

#=>x^2=(5^(1/4) + 5^(-1/4))^2=5^(1/2)+5^(-1/2)+2xx5^(1/4)xx5^(-1/4#
and

,# y=5^(1/4) - 5^(-1/4)#

,#=> y^2=(5^(1/4) - 5^(-1/4))^2=5^(1/2)+5^(-1/2)-2xx5^(1/4)xx5^(-1/4#

So

#x^2+y^2=2(5^(1/2)+5^(-1/2))#

Hence

#5(x^2+y^2)^2=5xx2(5^(1/2)+5^(-1/2))^2#

#=5xx2^2(sqrt5+1/sqrt5)^2#

#=5xx2^2((5+1)/sqrt5)^2#

#=cancel5xx2^2xx36/cancel5=144#

Answer 2 · 2017-03-10T12:01:33

See below.

#x=a+1/a# and #y=a-1/a# so

#xy=a^2-1/a^2# but

#(x+y)^2=x^2+y^2+2xy# so

#x^2+y^2=(2a)^2-2(a^2-1/a^2) = 2(a^2+1/a^2)# and also

#(x^2+y^2)^2=4(a^4+2+1/a^4)#

substituting #a^4=5# we get at

#5(x^2+y^2)^2=5 cdot 4(5+2+1/5)=144#