An algebra problem by atharva bhoirekar

$Given\quad that\\ \qquad \qquad \frac { x+1 }{ x } =2\\ =>\quad \quad \quad \quad x+1=2x\quad or\quad \boxed { x=1 } \\ To\quad Find\\ \qquad \qquad \frac { { x }^{ 2 }+1 }{ { x }^{ 2 } } \quad =\quad \frac { 1+1 }{ 1 } \qquad \because \quad \boxed { x=1 } \\ So,\quad Answer\quad is\quad \boxed { 2 }$

x+1 = 2x

Then, subtract x from both sides to get:

x=1

Plug in 1 for x in x^2+1/x^2 to get

1^2+1/1^2

which reduces to 1+1/1 which equals to 2

x+1/x=2

Squaring both sides

x^2 + 2(x)(1/x) + 1/x^2=4

Simplifying:

x^2 + 1/x^2 = 2

Thus, the answer is 2.

Other solution:

x^n + 1/x^n = 2 cos nA

Then substitute and you can arrive with the same answer which is 2.