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

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• 1 year ago

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

• 1 year ago

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.

• Ian H
Lv 7
1 year ago

[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

• 1 year ago

(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