If $a^2 - 4a + 1 = 0$, then the value of $ a^2 + a + \frac{1}{a} +\frac{1}{a^2}$ is :

18

We know that,

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If $a^2 - 4a + 1 = 0$,

then the value of $ a^2 + a + \frac{1}{a} +\frac{1}{a^2}$= ?

If $a^2 - 4a + 1 = 0$

Divide by a on both the sides of the equation we get,

a + \(\frac{1}{a}\) = 4

then, a² + \(\frac{1}{a^2}\) = 4² – 2 = 14

Put these values in the required equation,

$ a^2 + a + \frac{1}{a} +\frac{1}{a^2}$= 14 + 4 = 18