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If $\sqrt{x} -\frac{1}{\sqrt{x}}=\sqrt{3}$, then what is the value of $x^4 +\frac{1}{x^4}$? |
531 7 623 527 |
527 |
If x - \(\frac{1}{x}\) = n then, $x^2 +\frac{1}{x^2}$ = n2 + 2 If $\sqrt{x} -\frac{1}{\sqrt{x}}=\sqrt{3}$ x + \(\frac{1}{x}\) = (\(\sqrt {3}\))2 + 2 = 5 $x^2 +\frac{1}{x^2}$ = 52 - 2 = 23 $x^4 +\frac{1}{x^4}$ = 232 - 2 = 527 |
