If $x^2-2 \sqrt{5} x+1=0$, then what is the value of $x^5+\frac{1}{x^5} ?$
Answer & explanation
Correct answer: option 3
If $x^2-2 \sqrt{5} x+1=0$,
then what is the value of $x^5+\frac{1}{x^5}$ = ?
x5 + $\frac{1}{x^5}$ = (x2 + $\frac{1}{x^2}$) × (x3 + $\frac{1}{x^3}$) – (x + $\frac{1}{x}$)
If $K+\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 – 2
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
If $x^2-2 \sqrt{5} x+1=0$,
then, divide equation by x on both sides,
x + \(\frac{1}{x}\) = $2\sqrt{5}$
$x^3 +\frac{1}{x^3}$ = ( $2\sqrt{5}$)3 - 3 × $2\sqrt{5}$ = $34\sqrt{5}$
$x^2 +\frac{1}{x^2}$ = ($2\sqrt{5}$)2 – 2 = 18
x5 + $\frac{1}{x^5}$ = 18 × $34\sqrt{5}$ - $2\sqrt{5}$
x5 + $\frac{1}{x^5}$ = $612\sqrt{5}$ - $2\sqrt{5}$ = $610 \sqrt{5}$