If 12 x^2-21 x+1=0, then what is the value of 9 x^2+≤ft(16 x^2)^-1 ?

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

then, $K^2+\frac{1}{K^2}$ = n² – 2 × k × \(\frac{1}{k}\)

If $12 x^2-21 x+1=0$

then what is the value of $9 x^2+\left(16 x^2\right)^{-1}$

Divide the equation If $12 x^2-21 x+1=0$ by 4x on both the sides to get the desired format of the equation,

3x + \(\frac{1}{4x}\) = \(\frac{21}{4}\) 

$9 x^2+\left(16 x^2\right)^{-1}$ = (\(\frac{21}{4}\))² – 2 × 3x × \(\frac{1}{4x}\)

$9 x^2+\left(16 x^2\right)^{-1}$ = \(\frac{441}{16}\) - \(\frac{3}{2}\) 

$9 x^2+\left(16 x^2\right)^{-1}$ = \(\frac{441 - 24}{16}\) = $\frac{417}{16}$