\begin{tabular}{l} 03. Si: \( x-x^{-1}=2 \) \\ Halle: \( M=\sqrt{x^{4}+x^{-4}+15} \) \\ \begin{tabular}{lll} A) 7 & B) 9 & C) 16 \\ D) 25 & E) 49 & \\ \hline\end{tabular} \\ \hline\end{tabular}
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First, we start with the equation \( x - x^{-1} = 2 \). To solve for \( x \), let's set \( y = x - x^{-1} \). This gives us the quadratic equation: \[ x^2 - 2x - 1 = 0. \] Using the quadratic formula, we get: \[ x = \frac{2 \pm \sqrt{(2)^2 + 4}}{2} = 1 \pm \sqrt{2}. \] Next, we need to compute \( M = \sqrt{x^4 + x^{-4} + 15} \). We can find \( x^4 + x^{-4} \) by first getting \( x^2 + x^{-2} \): \[ x^2 + x^{-2} = (x - x^{-1})^2 + 2 = 2^2 + 2 = 4 + 2 = 6. \] Now, we can find \( x^4 + x^{-4} \): \[ x^4 + x^{-4} = (x^2 + x^{-2})^2 - 2 = 6^2 - 2 = 36 - 2 = 34. \] Finally, substituting this back into the formula for \( M \): \[ M = \sqrt{34 + 15} = \sqrt{49} = 7. \] The answer is **A) 7**.