If $x - 3=\frac{1}{2x},$ then what is the value of $(x^4 +\frac{1}{16x^4})$ ?

$99\frac{1}{2}$

We know that,

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

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

If $x - 3=\frac{1}{2x},$

then what is the value of $(x^4 +\frac{1}{16x^4})$ = ?

If $x - 3=\frac{1}{2x},$

then, If $x - \frac{1}{2x} =3$

so, $x^2 + \frac{1}{4x^2}$ = 3² + 2 × 1 × \(\frac{1}{2}\) = 10

then, $(x^4 +\frac{1}{16x^4})$ = 10² - 2 × 1 × \(\frac{1}{4}\) = $99\frac{1}{2}$