Questions
A. Rationalise the denominators of the following expressions, without the use of a catculator:

Ask by Schwartz Pena. in South Africa

Jan 24,2025

Question

Questions
A. Rationalise the denominators of the following expressions, without the use of a catculator:

Ask by Schwartz Pena. in South Africa

Jan 24,2025

UpStudy AI · Accepted Answer

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.
To rationalize the denominators: 1. For $$ \frac{2}{4 + \sqrt{2}} $$, multiply numerator and denominator by $$ 4 - \sqrt{2} $$: $$ \frac{2(4 - \sqrt{2})}{(4 + \sqrt{2})(4 - \sqrt{2})} = \frac{8 - 2\sqrt{2}}{14} = \frac{4 - \sqrt{2}}{7} $$ 2. For $$ \frac{6}{3 - \sqrt{5}} $$, multiply numerator and denominator by $$ 3 + \sqrt{5} $$: $$ \frac{6(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{18 + 6\sqrt{5}}{4} = \frac{9 + 3\sqrt{5}}{2} $$ Final answers: $$ \frac{4 - \sqrt{2}}{7} \quad \text{and} \quad \frac{9 + 3\sqrt{5}}{2} $$

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