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Home » Divide by $ d^2 $: let $ x = \fracAd $, then: - Esdistancia

Divide by $ d^2 $: let $ x = \fracAd $, then: - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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📌 äºæ¬¡æ¹ç¨å¼ \(2x^2 - 4x - 6 = 0\) ãä¸ãããã¦ãã¾ããäºæ¬¡æ¹ç¨å¼ã®è§£ãæ±ãã¦ãã ããã

📌 x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a}

📌 \(2x^2 - 4x - 6 = 0\) ã§ã¯ã\(a = 2\)ã\(b = -4\)ã\(c = -6\) ã§ãã

📌 b^2 - 4ac = (-4)^2 - 4 imes 2 imes (-6) = 16 + 48 = 64

📌 x = rac{-(-4) \pm \sqrt{64}}{2 imes 2} = rac{4 \pm 8}{4}

📌 x_1 = rac{4 + 8}{4} = 3, \quad x_2 = rac{4 - 8}{4} = -1

📌 é¢æ° \(f(x) = 3x^3 - 5x^2 + 2x - 7\) ã®å°é¢æ°ãæ±ãã\(x = 2\) ã§ã®å¤ãè©ä¾¡ãã¦ãã ããã

📌 f'(x) = rac{d}{dx}(3x^3 - 5x^2 + 2x - 7) = 9x^2 - 10x + 2

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But over hour 1: $ f(1) = k/1 = 48 $, so $ k = 48 $.
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For $ n=2 $: $ E_2 = -13.6 / 2^2 = -13.6 / 4 = -3.4 $ eV.
For $ n=3 $: $ E_3 = -13.6 / 3^2 = -13.6 / 9 \approx -1.511 $ eV.
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