If a+(1/a)=4 provide answer for a^4+(1/a^4)=?

Answer 1

Solution:-
a+(1/a)=4……………..eq(1)
squaring both said on eq(1)
(a+(1/a))^2=4^2
Now by solving (a+b)^2 at LHS we get
a^2+2(a*(1/a))+(1/a)^2=16
a^2+(1/a)^2=16-2
a^2+(1/a)^2=14…………..eq(2)
again squaring both said
then,
(a^2+(1/a^2))=14^2
a^4+2(a^2*(1/a^2))+(1/a^2)^2=196
a^4+(1/a^4)=196-2
a^4+(1/a^4)=194
Answer is 194.

Answer 2

a+(1/a)=4
a^2+1=4a
a^2-4a=-1
a^2-4a+4 = 3
(a-2)^2 = 3
a=2+√3 or a=2−√3

If a= 2+√3 then a^4+(1/a^4) = (2+√3)^4 + (1/(2+√3)^4)) = 97+56√3 + 1/(97+56√3) = 97+56√3 + (97-56√3)/(9409-9408) = 97+56√3 + (97-56√3)/(1) = 194.

If a=2−√3 then a^4+(1/a^4) = (2−√3)^4 + (1/(2−√3)^4)) = 97-56√3 + 1/(97-56√3) = 97-56√3 + (97+56√3)/(9409-9408) = 97-56√3 + (97+56√3)/(1) = 194.

Answer 3

[a + (1/a)]^4 = a^4 + 4a^2 + 6 + 4/a^2 + 1/a^4 = 256
4*[a + (1/a)]^2 = .... 4a^2 + 8 + 4/a^2 = 64
a^4 + 1/a^4 = 256 – 64 + 2 = 194

Answer 4

(x + y)^4 = x^4 + 4x^3 * y + 6x^2 * y^2 + 4x * y^3 + y^4
a + 1/a = 4
(a + 1/a)^4 = 4^4 = 256

256 = a^4 + (1/a)^4 + 4 * a^3 * (1/a) + 6 * a^2 * (1/a)^2 + 4 * a * (1/a)^3
256 = a^4 + (1/a)^4 + 4a^2 + 6 + 4/a^2
250 = a^4 + (1/a)^4 + 4 * (a^2 + (1/a)^2)

(a + (1/a))^2 = 4^2 = 16 = a^2 + 2 * a/a + (1/a)^2 = a^2 + (1/a)^2 + 2

250 = a^4 + (1/a)^4 + 4 * (a^2 + 2 + (1/a)^2 - 2)
250 = a^4 + (1/a)^4 + 4 * (16 - 2)
250 = a^4 + (1/a)^4 + 4 * 14
250 = a^4 + (1/a)^4 + 56
194 = a^4 + (1/a)^4

Answer 5

a + (1/a) = 4
a^4 + (1/a^4) = 194