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