If $ x + \frac{1}{x} = \sqrt{7}$, then what is the value of $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$?
Answer & explanation
Correct answer: option 4
If $ x + \frac{1}{x} = \sqrt{7}$,
then what is the value of $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$
We can write $(x^2 + 1) ÷ [ x^4 +(\frac{1}{x^2})]$ this equation as (x + \(\frac{1}{x}\)) ÷ [ x3 + \(\frac{1}{x^3}\)] by dividing x on both numerator and denominator.
We know that,
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
$x^3 +\frac{1}{x^3}$ = ($\sqrt{7}$)3 - 3 × $\sqrt{7}$ = $4\sqrt{7}$
Put these values in the required equation,
= $(x + \frac{1}{x})$ ÷ [ $x^3 +(\frac{1}{x^3})]$ = $\sqrt{7}$ ÷ $4\sqrt{7}$ = $\frac{1}{4}$