If #x + 1/x = 11# , then find the value of #x^4 + 1/x^4# ?

Question

19502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T06:44:23

#x^4+1/x^4 = 14159#

As per the question, we have

#x+1/x = 11#

#:.(x+1/x)^2=(11)^2# ... [Squaring both sides]

#:.x^2+1/x^2+2(x)(1/x)=121#

#:.x^2+1/x^2+2(cancelx)(1/cancelx)=121#

#:.x^2+1/x^2+2=121#

#:.x^2+1/x^2=121-2=119# ... (i)

Now, back to #x+1/x=11#

#(x+1/x)^4=(11)^4#

#:.x^4+1/x^4+4(x^3)(1/x)+6(x^2)(1/x^2)+4(x)(1/x^3)=(11)^4#

#:.x^4+1/x^4+4(x^2)+6+4(1/x^2)=14641#

#:.x^4+1/x^4+4(x^2)+4(1/x^2)=14641-6#

#:.x^4+1/x^4+4(x^2+1/x^2)=14635#

#:.x^4+1/x^4+4(119)=14635# ... [Substituting the value of #x^2+1/x^2# from (i)]

#:.x^4+1/x^4+476=14635#

#:.x^4+1/x^4=14635-476#

#:.x^4+1/x^4=14159#

Hence, the answer.

Note: If you want to know more about the identity #(a+b)^4# or any other, go to THIS SITE.

Answer 2 · 2018-06-09T07:13:04

#14159#

We use that
#(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4#
so we get
#(x+1/x)^4=x^4+1/x^4+4(x^2+1/x^2)=11^4-6#
and
#x^2+1/x^2=11^2-2#
so we get

#x^4+1/x^4=11^4-6-4(11^2-2)=14159#

Answer 3 · 2018-06-09T14:52:04

#x^2+1/x^2=(x+1/x)^2-2*x*1/x#

#=>x^2+1/x^2=11^2-2=119#

Now #x^4+1/x^4=(x^2+1/x^2)^2-2*x^2*1/x^2#

#=>x^4+1/x^4=119^2-2=14159#