Algebra Frenzy - Part 4

Let us assign $x ^ 2 -\frac {1}{x ^ 2} = a$

( $x + \frac {1}{x}$ ) ( $x - \frac {1}{x}$ ) $= a$

Substituting in from given equation: $3$ ( $x - \frac {1}{x}$ ) $= a$ and therefore ( $x - \frac {1}{x}$ ) $= \frac {a}{3}$

( $x - \frac {1}{x}) ^ 2$ $= x ^ 2 + \frac{1}{x ^ 2} - 2$ $= \frac {a ^ 2}{9}$

$x ^ 2 + \frac{1}{x ^ 2} + 2$ $= \frac {a ^ 2}{9} + 4$

Square rooting both sides we get: $x + \frac {1}{x} = \sqrt{\frac {a ^ 2}{9} + 4}$

Substituting in from given equation: $3 = \sqrt{\frac {a ^ 2}{9} + 4}$

Rearrange for a $9 = \frac {a ^ 2}{9} + 4$

$5 = \frac {a ^ 2}{9}$ and $a ^ 2 = 45$

$a = \boxed{\pm 3 {\sqrt{5}}}$