If $(x^2 +\frac{1}{49x^2}) = 15\frac{5}{7}$, then what is the value of $(x + \frac{1}{7x})$ ?

±4

We know that,

If , x² + \(\frac{1}{x^2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2 \times x \times \frac{1}{x}}\)

$(x^2 +\frac{1}{49x^2}) = 15\frac{5}{7}$, = $\frac{110}{7}$

then what is the value of $(x + \frac{1}{7x})$ = \(\sqrt {\frac{110}{7} + \frac{2}{7}}\) = \(\sqrt {\frac{112}{7}}\)

$(x + \frac{1}{7x})$ = \(\sqrt {16}\) = ±4