Questions A. Rationalise the denominators of the following expressions, without the use of a catculator: \( \begin{array}{ll}\text { 3. } \frac{2}{4+\sqrt{2}} & \text { 4. } \frac{6}{3-\sqrt{5}}\end{array} \)

To rationalize the denominators of the given expressions without using a calculator, we'll use the method of multiplying by the **conjugate** of the denominator. The conjugate of a binomial \( a + b \) is \( a - b \), and vice versa. This method helps eliminate the square roots in the denominator. Let's tackle each expression step-by-step. --- ### **3. \( \frac{2}{4 + \sqrt{2}} \)** **Step 1: Identify the Conjugate** The denominator is \( 4 + \sqrt{2} \). Its conjugate is \( 4 - \sqrt{2} \). **Step 2: Multiply Numerator and Denominator by the Conjugate** \[ \frac{2}{4 + \sqrt{2}} \times \frac{4 - \sqrt{2}}{4 - \sqrt{2}} = \frac{2(4 - \sqrt{2})}{(4 + \sqrt{2})(4 - \sqrt{2})} \] **Step 3: Simplify the Denominator** \[ (4 + \sqrt{2})(4 - \sqrt{2}) = 4^2 - (\sqrt{2})^2 = 16 - 2 = 14 \] **Step 4: Expand the Numerator** \[ 2(4 - \sqrt{2}) = 8 - 2\sqrt{2} \] **Step 5: Combine the Results** \[ \frac{8 - 2\sqrt{2}}{14} = \frac{8}{14} - \frac{2\sqrt{2}}{14} = \frac{4}{7} - \frac{\sqrt{2}}{7} = \frac{4 - \sqrt{2}}{7} \] **Final Answer:** \[ \boxed{\dfrac{4 - \sqrt{2}}{7}} \] --- ### **4. \( \frac{6}{3 - \sqrt{5}} \)** **Step 1: Identify the Conjugate** The denominator is \( 3 - \sqrt{5} \). Its conjugate is \( 3 + \sqrt{5} \). **Step 2: Multiply Numerator and Denominator by the Conjugate** \[ \frac{6}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{6(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} \] **Step 3: Simplify the Denominator** \[ (3 - \sqrt{5})(3 + \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4 \] **Step 4: Expand the Numerator** \[ 6(3 + \sqrt{5}) = 18 + 6\sqrt{5} \] **Step 5: Combine the Results** \[ \frac{18 + 6\sqrt{5}}{4} = \frac{18}{4} + \frac{6\sqrt{5}}{4} = \frac{9}{2} + \frac{3\sqrt{5}}{2} = \frac{9 + 3\sqrt{5}}{2} \] **Final Answer:** \[ \boxed{\dfrac{9 + 3\sqrt{5}}{2}} \] --- These rationalized forms eliminate the radicals from the denominators, making the expressions simpler and more standardized.