If $20x^{2} — 30x + 1 = 0$, then what is the value of $25x^{2}+\frac{1}{16x^{2}}$

53$\frac{3}{4}$

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

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

If $20x^{2} — 30x + 1 = 0$,

then what is the value of $25x^{2}+\frac{1}{16x^{2}}$

Divide $20x^{2} — 30x + 1 = 0$ by 5x on both sides to get the desired format of the equation,

5x + \(\frac{1}{4x^2}\) = \(\frac{30}{4}\) = \(\frac{15}{2}\) 

So, $25x^{2}+\frac{1}{16x^{2}}$ = ( \(\frac{15}{2}\) )² – 2 × 5x × \(\frac{1}{4x}\)

$25x^{2}+\frac{1}{16x^{2}}$ =  \(\frac{225}{4}\)  - \(\frac{5}{2}\)

$25x^{2}+\frac{1}{16x^{2}}$ = \(\frac{225 - 10}{4}\) = \(\frac{215}{4}\)

$25x^{2}+\frac{1}{16x^{2}}$ = 53$\frac{3}{4}$