Is There a Winning Lawyer For Drug Charges in Your City?

Is There a Winning Lawyer For Drug Charges in Your City?

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Let $ x = \frac{a}{b} $, then $ x + \frac{1}{x} = 5 \Rightarrow x^2 + \frac{1}{x^2} = 23 $, so:x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{Not helpful}Instead, use:\frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1} = \frac{(x^2 + 1)^2 + (x^2 - 1)^2}{(x^2 - 1)(x^2 + 1)} = \frac{2x^4 + 2}{x^4 - 1} = 2 \cdot \frac{x^4 + 1}{x^4 - 1}Let’s compute $ x^4 + 1 $, $ x^4 - 1 $:We know $ x^2 + \frac{1}{x^2} = 23 \Rightarrow x^4 + \frac{1}{x^4} = 527 $, so:x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{Still not helpful}Let’s try plugging values:Let $ x = \frac{5 + \sqrt{21}}{2} $, then $ x + \frac{1}{x} = 5 $. Compute numerically:x = \frac{5 + \sqrt{21}}{2} \approx \frac{5 + 4.583}{2} = \frac{9.583}{2} \approx 4.7915