Answer
**Question 1 Solutions:**
1. **1.1.1**
- If \( x = 2 \), then \( k = 6 \).
2. **1.1.2**
- If \( k = 2 \), then \( x = 1 \) and \( x = 6 \).
3. **1.2.1**
- The solutions for \( x \) are approximately \( 3.58 \) and \( 0.42 \).
4. **1.2.2**
- The solution for \( x \) is \( 8 \).
5. **1.2.3**
- All real numbers \( x \) where \( x < 5 \).
6. **1.3**
- The simultaneous solutions are \( (6, 0) \) and \( (2, 1) \).
Solution
Sure, let's solve each part of **Question 1** step-by-step.
---
### **Question 1.1**
**Given:** \( 2k = (x - 5)(x - k) \)
#### **1.1.1. Determine \( k \) if \( x = 2 \)**
**Step 1:** Substitute \( x = 2 \) into the equation.
\[
2k = (2 - 5)(2 - k)
\]
**Step 2:** Simplify the equation.
\[
2k = (-3)(2 - k) \\
2k = -6 + 3k
\]
**Step 3:** Solve for \( k \).
\[
2k - 3k = -6 \\
- k = -6 \\
k = 6
\]
**Answer:** \( k = 6 \)
---
#### **1.1.2. Determine \( x \) if \( k = 2 \)**
**Step 1:** Substitute \( k = 2 \) into the equation.
\[
2 \times 2 = (x - 5)(x - 2) \\
4 = (x - 5)(x - 2)
\]
**Step 2:** Expand the right side.
\[
4 = x^2 - 2x - 5x + 10 \\
4 = x^2 - 7x + 10
\]
**Step 3:** Rearrange the equation to standard quadratic form.
\[
x^2 - 7x + 6 = 0
\]
**Step 4:** Solve the quadratic equation using the quadratic formula.
\[
x = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 1 \times 6}}{2 \times 1} \\
x = \frac{7 \pm \sqrt{49 - 24}}{2} \\
x = \frac{7 \pm \sqrt{25}}{2} \\
x = \frac{7 \pm 5}{2}
\]
**Solutions:**
\[
x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{2}{2} = 1
\]
**Answer:** \( x = 1 \) and \( x = 6 \)
---
### **Question 1.2**
#### **1.2.1. Solve for \( x \): \( 2x^2 + 3 = 8x \) (correct to TWO decimal places)**
**Step 1:** Rearrange the equation to standard quadratic form.
\[
2x^2 - 8x + 3 = 0
\]
**Step 2:** Apply the quadratic formula.
\[
x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 2 \times 3}}{2 \times 2} \\
x = \frac{8 \pm \sqrt{64 - 24}}{4} \\
x = \frac{8 \pm \sqrt{40}}{4} \\
x = \frac{8 \pm 2\sqrt{10}}{4} \\
x = \frac{4 \pm \sqrt{10}}{2}
\]
**Step 3:** Calculate the numerical values.
\[
\sqrt{10} \approx 3.1623 \\
x \approx \frac{4 + 3.1623}{2} \approx 3.58 \\
x \approx \frac{4 - 3.1623}{2} \approx 0.42
\]
**Answer:** \( x \approx 3.58 \) and \( x \approx 0.42 \)
---
#### **1.2.2. Solve for \( x \): \( \sqrt{2(x + 10)} - 10 = x - 12 \)**
**Step 1:** Isolate the square root.
\[
\sqrt{2(x + 10)} = x - 12 + 10 \\
\sqrt{2(x + 10)} = x - 2
\]
**Step 2:** Determine the domain.
- The expression inside the square root must be non-negative:
\[
2(x + 10) \geq 0 \Rightarrow x \geq -10
\]
- The right side must be non-negative:
\[
x - 2 \geq 0 \Rightarrow x \geq 2
\]
Thus, \( x \geq 2 \).
**Step 3:** Square both sides to eliminate the square root.
\[
2(x + 10) = (x - 2)^2 \\
2x + 20 = x^2 - 4x + 4
\]
**Step 4:** Rearrange to form a quadratic equation.
\[
x^2 - 6x - 16 = 0
\]
**Step 5:** Solve the quadratic equation.
\[
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times (-16)}}{2 \times 1} \\
x = \frac{6 \pm \sqrt{36 + 64}}{2} \\
x = \frac{6 \pm \sqrt{100}}{2} \\
x = \frac{6 \pm 10}{2}
\]
**Solutions:**
\[
x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-4}{2} = -2
\]
**Step 6:** Check the solutions against the domain \( x \geq 2 \).
- \( x = 8 \) is valid.
- \( x = -2 \) is rejected.
**Answer:** \( x = 8 \)
---
#### **1.2.3. Solve the inequality: \( 3^{x}(x - 5) < 0 \)**
**Step 1:** Analyze the factors.
- \( 3^x \) is always positive for all real \( x \) since the base \( 3 > 1 \).
- \( x - 5 \) changes sign at \( x = 5 \).
**Step 2:** Determine where the product is negative.
Since \( 3^x > 0 \), the sign of the product \( 3^x(x - 5) \) depends solely on \( x - 5 \).
\[
3^x(x - 5) < 0 \quad \Rightarrow \quad x - 5 < 0 \quad \Rightarrow \quad x < 5
\]
**Answer:** All real numbers \( x \) such that \( x < 5 \)
---
### **Question 1.3**
**Solve the following equations simultaneously:**
\[
\begin{cases}
\sqrt{3^x} \cdot 9^y = 27 \\
x + 4y^2 = 6
\end{cases}
\]
**Step 1:** Simplify the first equation by expressing all terms with base 3.
\[
\sqrt{3^x} = 3^{x/2} \quad \text{and} \quad 9^y = (3^2)^y = 3^{2y}
\]
\[
3^{x/2} \cdot 3^{2y} = 3^3 \\
3^{x/2 + 2y} = 3^3
\]
**Step 2:** Equate the exponents since the bases are the same.
\[
\frac{x}{2} + 2y = 3
\]
Multiply both sides by 2:
\[
x + 4y = 6 \quad \text{(Equation A)}
\]
**Step 3:** Use Equation A and the second original equation to solve for \( x \) and \( y \).
\[
\begin{cases}
x + 4y = 6 \quad \text{(Equation A)} \\
x + 4y^2 = 6 \quad \text{(Equation B)}
\end{cases}
\]
**Step 4:** Subtract Equation A from Equation B.
\[
(x + 4y^2) - (x + 4y) = 0 \\
4y^2 - 4y = 0 \\
4y(y - 1) = 0
\]
**Solutions for \( y \):**
\[
y = 0 \quad \text{or} \quad y = 1
\]
**Step 5:** Find corresponding \( x \) values.
- **Case 1:** \( y = 0 \)
\[
x + 4(0) = 6 \Rightarrow x = 6
\]
- **Case 2:** \( y = 1 \)
\[
x + 4(1) = 6 \Rightarrow x = 2
\]
**Solutions:**
\[
(x, y) = (6, 0) \quad \text{and} \quad (2, 1)
\]
**Answer:** The simultaneous solutions are \( (x, y) = (6,\ 0) \) and \( (2,\ 1) \)
---
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