If $a^2 - 4a + 1 = 0$, then the value of $ a^2 + a + \frac{1}{a} +\frac{1}{a^2}$ is :
Answer & explanation
Correct answer: option 3
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