QUESTION 4 Round off the following numbers to the decimal places indicated: 4.1.1 4,3768534 ( 4 decimal places \( ) \) 4.1.2 \( \sqrt{7,24357}(1 \) decimal place \( ) \) Given: \( p=\sqrt{b^{2}+4 \mathrm{ac}} \) 4.2.1 Determine the value of \( p \) if \( a=2 \); \( b=6 \); \( c=2 \) (Leave your answer in the simplest surd form) 4.2.2 State whether \( p \) is rational or irrational. 4.2.3 Between which TWO consecutive integers does \( p \) lie? Write the following recurring decimals in fraction form: 4.3.1 \( \begin{array}{l}\text { 4.3 } \\ \text { 4. }\end{array} \)

Certainly! Let's address each part of **QUESTION 4** step by step. --- ### **4.1.1 Rounding 4.3768534 to 4 Decimal Places** **Number:** 4.3768534 **Decimal Places to Round:** 4 **Steps:** 1. Identify the fourth decimal place: **4.3768**. 2. Look at the fifth decimal place to decide whether to round up or stay: The fifth digit is **5**. 3. Since the fifth digit is **5** or more, round up the fourth decimal place by 1. **Rounded Number:** **4.3769** --- ### **4.1.2 Rounding √7.24357 to 1 Decimal Place** **Expression:** √7.24357 **Decimal Places to Round:** 1 **Steps:** 1. Calculate the square root of 7.24357. - √7 ≈ 2.6458 - √7.24357 ≈ **2.692** (approximate value). 2. Round to 1 decimal place: - The first decimal place is **6**. - The second decimal place is **9** (since it's ≥5, round up). **Rounded Value:** **2.7** --- ### **4.2.1 Determining the Value of \( p = \sqrt{b^2 + 4ac} \) with \( a = 2 \), \( b = 6 \), \( c = 2 \)** **Given:** \( a = 2 \) \( b = 6 \) \( c = 2 \) **Calculation:** \[ p = \sqrt{b^2 + 4ac} = \sqrt{6^2 + 4 \times 2 \times 2} = \sqrt{36 + 16} = \sqrt{52} \] Simplify \( \sqrt{52} \): \[ \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \] **Value of \( p \):** \( p = 2\sqrt{13} \) --- ### **4.2.2 Determining Whether \( p \) is Rational or Irrational** **Expression:** \( p = 2\sqrt{13} \) **Analysis:** - \( \sqrt{13} \) is an **irrational** number because 13 is not a perfect square. - Multiplying an irrational number by a rational number (2) still results in an **irrational** number. **Conclusion:** \( p \) is **Irrational**. --- ### **4.2.3 Identifying the Two Consecutive Integers Between Which \( p \) Lies** **Expression:** \( p = 2\sqrt{13} \) **Steps:** 1. Approximate \( \sqrt{13} \): - \( \sqrt{9} = 3 \) - \( \sqrt{16} = 4 \) - So, \( \sqrt{13} \) is between **3** and **4**. - More precisely, \( \sqrt{13} \approx 3.6055 \). 2. Calculate \( p \): \[ p = 2 \times 3.6055 \approx 7.211 \] 3. Identify the consecutive integers: - **7** and **8**. **Conclusion:** \( p \) lies between **7** and **8**. --- ### **4.3.1 Writing Recurring Decimals in Fraction Form** *It appears there might be some formatting issues with the recurring decimals you intended to provide in part 4.3.1. However, I can demonstrate how to convert a recurring decimal to a fraction using an example.* **Example:** Convert \( 0.\overline{6} \) to a fraction. **Steps:** 1. Let \( x = 0.\overline{6} \). 2. Multiply both sides by 10 to shift the decimal point: \[ 10x = 6.\overline{6} \] 3. Subtract the original equation from this new equation: \[ 10x - x = 6.\overline{6} - 0.\overline{6} \] \[ 9x = 6 \] 4. Solve for \( x \): \[ x = \frac{6}{9} = \frac{2}{3} \] **Conclusion:** \( 0.\overline{6} = \frac{2}{3} \) *If you provide the specific recurring decimals for parts 4.3.1 and 4.3.2, I can offer precise conversions.* --- Feel free to provide the specific recurring decimals for **4.3.1** and **4.3.2**, and I'll be happy to assist further!