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\( \frac { ( a - \varepsilon ) ^ { 4 } } { b ^ { 4 } } - \frac { 3 ^ { * } ( a - \varepsilon ) ^ { 2 } } { b ^ { 2 } } + 1 = 0 \)

Ask by Lawson Rojas. in Greece
Jan 20,2025

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Tutor-Verified Answer

Answer

Οι λύσεις της εξίσωσης είναι: \[ \varepsilon = a \pm b \sqrt{\frac{3 + \sqrt{5}}{2}}, \quad \varepsilon = a \pm b \sqrt{\frac{3 - \sqrt{5}}{2}} \]

Solution

Λύση της εξίσωσης: \[ \frac{(a - \varepsilon)^4}{b^4} - \frac{3(a - \varepsilon)^2}{b^2} + 1 = 0 \] **Βήμα 1:** Κάνουμε την αντικατάσταση \( x = \frac{(a - \varepsilon)^2}{b^2} \). Η εξίσωση γίνεται: \[ x^2 - 3x + 1 = 0 \] **Βήμα 2:** Λύνουμε την δευτεροβάθμια εξίσωση για το \( x \): \[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \] **Βήμα 3:** Επαναφέρουμε την αντικατάσταση για \( x \): \[ \frac{(a - \varepsilon)^2}{b^2} = \frac{3 \pm \sqrt{5}}{2} \] **Βήμα 4:** Λύνουμε για \( (a - \varepsilon) \): \[ a - \varepsilon = \pm b \sqrt{\frac{3 \pm \sqrt{5}}{2}} \] **Βήμα 5:** Λύνοντας για το \( \varepsilon \), παίρνουμε τις τέσσερις δυνατές λύσεις: \[ \varepsilon = a \pm b \sqrt{\frac{3 + \sqrt{5}}{2}} \quad \text{και} \quad \varepsilon = a \pm b \sqrt{\frac{3 - \sqrt{5}}{2}} \] Άρα, οι λύσεις της εξίσωσης είναι: \[ \varepsilon = a + b \sqrt{\frac{3 + \sqrt{5}}{2}}, \quad \varepsilon = a - b \sqrt{\frac{3 + \sqrt{5}}{2}}, \] \[ \varepsilon = a + b \sqrt{\frac{3 - \sqrt{5}}{2}}, \quad \varepsilon = a - b \sqrt{\frac{3 - \sqrt{5}}{2}} \]

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Mind Expander

This equation resembles a polynomial in terms of \( x = \frac{(a - \varepsilon)}{b} \). When you rewrite it, it becomes \( x^4 - 3x^2 + 1 = 0 \), which is a quadratic equation in terms of \( x^2 \). You can set \( y = x^2 \) to get the factorable form as \( y^2 - 3y + 1 = 0 \). Using the quadratic formula gives you solutions for \( y \), and from there, you can find the corresponding \( x \) values, which in turn leads you to the values of \( a \) and \( \varepsilon \). Once you have obtained the roots, don’t forget to verify whether they fit the original equation and check if there are any restrictions on \( a \) or \( \varepsilon \) that might affect the solutions. Common mistakes include overlooking these conditions or forgetting to check for extraneous solutions that arise from squaring both sides or simplifying.

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