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# help algebra

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Let a and b with a>b>0 be real numbers satisfying a^2+b^2=4ab. Find a/b - b/a.

May 28, 2022

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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.

May 28, 2022