If $b^2 - 4b - 1 = 0$, the find the value of $b^2 + \frac{1}{b^2} + 3b - \frac{3}{b}$.

30

We know that,

If x - \(\frac{1}{x}\)  = n

then, x² + \(\frac{1}{x^2}\)  = \(\sqrt {n^2 + 2}\)

If $b^2 - 4b - 1 = 0$,

The find the value of $b^2 + \frac{1}{b^2} + 3b - \frac{3}{b}$ = ?

Divide on both the sides of If $b^2 - 4b - 1 = 0$ by b we get,

b - \(\frac{1}{b}\) = 4

The value of $b^2 + \frac{1}{b^2}$ =  \(\sqrt {4^2 + 2}\) = 18

Put the value of these into the required equation,

$18 + 3(b - \frac{1}{b})$ = 18 + 3(4) = 30