Pregunta

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

Solución de inteligencia artificial de Upstudy

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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 \[ \frac{1}{4}. \]

Solución

**1.2.1** We simplify \[ (7+\sqrt{2})(6-\sqrt{2}) \] by expanding the product: \[ (7+\sqrt{2})(6-\sqrt{2}) = 7\cdot 6 + 7\cdot (-\sqrt{2}) + \sqrt{2}\cdot 6 + \sqrt{2}\cdot (-\sqrt{2}). \] Calculating each term: \[ 7\cdot 6 = 42,\quad 7\cdot (-\sqrt{2}) = -7\sqrt{2},\quad \sqrt{2}\cdot 6 = 6\sqrt{2},\quad \sqrt{2}\cdot (-\sqrt{2}) = -2. \] Combine like terms: \[ 42 - 7\sqrt{2} + 6\sqrt{2} - 2 = (42-2) + (-7\sqrt{2}+6\sqrt{2}) = 40 - \sqrt{2}. \] Thus, the simplified expression is \[ 40 - \sqrt{2}. \] --- **1.2.2** Determine the nature of the roots for \[ z^2 = 3. \] We rewrite the equation as \[ z^2 - 3 = 0. \] Taking the square root of both sides gives: \[ z = \sqrt{3} \quad \text{or} \quad z = -\sqrt{3}. \] The two distinct solutions are real numbers. Therefore, the equation has two distinct real roots. --- **1.2.3** Simplify \[ \frac{45^{-x+1} \cdot 5^{x-1} \cdot 81^{-1}}{4^{x+2} \cdot 36^{-x-1}}. \] *Step 1. Rewrite numbers in terms of their prime factorizations:* \[ 45 = 3^2 \cdot 5,\quad 81 = 3^4,\quad 4 = 2^2,\quad 36 = 2^2 \cdot 3^2. \] *Step 2. Rewrite the numerator:* \[ 45^{-x+1} = (3^2 \cdot 5)^{-x+1} = 3^{2(-x+1)} \cdot 5^{-x+1}, \] \[ 81^{-1} = 3^{-4}. \] Thus, the numerator is: \[ 3^{2(-x+1)} \cdot 5^{-x+1} \cdot 5^{x-1} \cdot 3^{-4}. \] Note that \[ 5^{-x+1} \cdot 5^{x-1} = 5^0 = 1. \] So it simplifies to: \[ 3^{2(-x+1)-4} = 3^{-2x+2-4} = 3^{-2x-2}. \] *Step 3. Rewrite the denominator:* \[ 4^{x+2} = (2^2)^{x+2} = 2^{2(x+2)} = 2^{2x+4}, \] \[ 36^{-x-1} = (2^2 \cdot 3^2)^{-x-1} = 2^{2(-x-1)} \cdot 3^{2(-x-1)} = 2^{-2x-2} \cdot 3^{-2x-2}. \] Thus, the denominator becomes: \[ 2^{2x+4} \cdot 2^{-2x-2} \cdot 3^{-2x-2} = 2^{(2x+4-2x-2)} \cdot 3^{-2x-2} = 2^{2} \cdot 3^{-2x-2} = 4 \cdot 3^{-2x-2}. \] *Step 4. Combine numerator and denominator:* \[ \frac{3^{-2x-2}}{4 \cdot 3^{-2x-2}}. \] Since \(3^{-2x-2}\) cancels in the numerator and denominator, the expression simplifies to \[ \frac{1}{4}. \] Thus, the simplified expression is \[ \frac{1}{4}. \]

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Bonus Knowledge

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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