Let a and b with a>b>0 be real numbers satisfying a^2+b^2=4ab. Find a/b - b/a.
+9519 Divide by ab on both sides of the equation.
\(\dfrac ab + \dfrac ba = 4\)
Squaring gives
\(\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} + 2 = 16\\ \dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} =14\)
Let x = a/b - b/a. Then x^2 = a^2/b^2 + b^2/a^2 - 2 = 12.
Therefore, \(\dfrac ab - \dfrac ba = \pm 2 \sqrt 3\). Since a > b > 0, the negative root is rejected. \(\dfrac ab - \dfrac ba = 2\sqrt 3\).
Please try to fill in the missing details.
