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 \( \frac{45^{-x+1} \cdot 5^{x-1} .81^{-1}}{4^{x+2} \cdot 36^{-x-1}} \)

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