If $\sqrt{x} −\frac{1}{\sqrt{x}} = \sqrt{5}$, x ≠ 0, then what is the value of $\frac{(x^4+\frac{1}{x^2})}{(x^2+1)}$?

46

If $\sqrt{x} −\frac{1}{\sqrt{x}} = \sqrt{5}$

Then, $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{5^2 + 4}$ = 3

x + \(\frac{1}{x}\) = 3² - 2 = 7

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

$x^3 +\frac{1}{x^3}$ = 7³ - 3 × 7 = 322

Now according to the question we have to find = $\frac{(x^4+\frac{1}{x^2})}{(x^2+1)}$

Take x common from numerator and denominator then we left with = $\frac{(x^3+\frac{1}{x^3})}{(x+\frac{1}{x})}$

= \(\frac{322}{7}\) = 46