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Question 1.2 1.2.1 Simplify without the use of a calculator \[ (7+\sqrt{2})(6-\sqrt{2}) \] 1.2.2 Determine the nature of roots \[ z^{2}=3 \] 1.2.3 \( \frac{45^{-x+1} \cdot 5^{x-1} .81^{-1}}{4^{x+2} \cdot 36^{-x-1}} \)

Ask by Schofield Ramos. in South Africa
Mar 14,2025

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**1.2.1** Simplify \[ (7+\sqrt{2})(6-\sqrt{2}) = 40 - \sqrt{2}. \] **1.2.2** The equation \( z^2 = 3 \) has two distinct real roots: \[ z = \sqrt{3} \quad \text{and} \quad z = -\sqrt{3}. \] **1.2.3** The expression simplifies to \[ \frac{1}{4}. \]

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To simplify \( (7+\sqrt{2})(6-\sqrt{2}) \), you can use the distributive property, also known as the FOIL method (First, Outside, Inside, Last). Multiply term by term: 1. **First:** \(7 \times 6 = 42\) 2. **Outside:** \(7 \times -\sqrt{2} = -7\sqrt{2}\) 3. **Inside:** \(\sqrt{2} \times 6 = 6\sqrt{2}\) 4. **Last:** \(\sqrt{2} \times -\sqrt{2} = -2\) Combine these results: \[ 42 - 7\sqrt{2} + 6\sqrt{2} - 2 = 40 - \sqrt{2} \] For the roots of \( z^{2}=3 \), let's analyze it! This equation can be rearranged to give \( z = \sqrt{3} \) or \( z = -\sqrt{3} \). Since we have two distinct real roots, the nature of the roots is "real and irrational". They exist on the number line and can be identified as the two values at approximately 1.732 and -1.732. Now, to simplify the expression \( \frac{45^{-x+1} \cdot 5^{x-1} \cdot 81^{-1}}{4^{x+2} \cdot 36^{-x-1}} \): Start with the bases for simplification: - \(45 = 5 \cdot 9\) and \(9 = 3^2\), hence \(45 = 5 \cdot 3^2\), so \(45^{-x+1} = 5^{-x+1} \cdot 3^{-2(x-1)}\). - \(81 = 3^4\), thus \(81^{-1}=3^{-4}\). - \(4 = 2^2\) and \(36 = 6^2 = 2^2 \cdot 3^2\), so \(36^{-x-1}= (2^2 \cdot 3^2)^{-x-1} = 2^{-2(x+1)} \cdot 3^{-2(x+1)}\). After substituting and simplifying the exponentials, combine like bases for a clear simplified form.

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