Answer
- **1.3:** The equation \( 2(x - 3)^2 + 2 = 0 \) has no real roots; it has two complex roots.
- **1.4:** The values of \( p \) are \( 1 \) and \( -1 \).
Solution
Let's address each problem step by step.
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### **1.3 Discuss the Nature of the Roots of the Equation \( 2(x - 3)^2 + 2 = 0 \)**
**Given Equation:**
\[ 2(x - 3)^2 + 2 = 0 \]
**Step 1: Simplify the Equation**
\[
2(x - 3)^2 + 2 = 0 \\
\Rightarrow 2(x - 3)^2 = -2 \\
\Rightarrow (x - 3)^2 = -1
\]
**Step 2: Analyze the Solution**
\[
(x - 3)^2 = -1
\]
The square of any real number is always **non-negative**. Therefore, \((x - 3)^2 = -1\) has no real solutions because a square cannot be negative.
**Conclusion:**
- **No Real Roots:** The equation has no real solutions.
- **Complex Roots:** It has two distinct complex (imaginary) roots.
---
### **1.4 Determine the Value(s) of \( p \) if \( g(x) = -2x^2 - px + 3 \) Has a Maximum Value of \( 3\, \frac{1}{8} \)**
**Given Function:**
\[ g(x) = -2x^2 - px + 3 \]
**Objective:** Find the value(s) of \( p \) such that the maximum value of \( g(x) \) is \( 3\, \frac{1}{8} \) (which is \( \frac{25}{8} \) when expressed as an improper fraction).
**Step 1: Identify the Vertex of the Parabola**
Since the coefficient of \( x^2 \) is negative (\( -2 \)), the parabola opens downward, and the vertex represents the maximum point.
The vertex \( (h, k) \) of a parabola \( g(x) = ax^2 + bx + c \) is given by:
\[
h = -\frac{b}{2a} \\
k = g(h)
\]
**Here:**
\[
a = -2, \quad b = -p, \quad c = 3
\]
**Calculating \( h \):**
\[
h = -\frac{-p}{2 \times -2} = \frac{p}{-4} = -\frac{p}{4}
\]
**Calculating \( k \) (Maximum Value):**
\[
k = g(h) = -2\left(-\frac{p}{4}\right)^2 - p\left(-\frac{p}{4}\right) + 3 \\
= -2\left(\frac{p^2}{16}\right) + \frac{p^2}{4} + 3 \\
= -\frac{p^2}{8} + \frac{p^2}{4} + 3 \\
= \left(-\frac{1}{8} + \frac{2}{8}\right)p^2 + 3 \\
= \frac{1}{8}p^2 + 3
\]
**Set \( k \) Equal to the Given Maximum Value:**
\[
\frac{1}{8}p^2 + 3 = \frac{25}{8}
\]
**Step 2: Solve for \( p \)**
\[
\frac{1}{8}p^2 = \frac{25}{8} - 3 \\
\frac{1}{8}p^2 = \frac{25}{8} - \frac{24}{8} \\
\frac{1}{8}p^2 = \frac{1}{8} \\
p^2 = 1 \\
p = \pm 1
\]
**Conclusion:**
- The values of \( p \) that satisfy the condition are:
\[ p = 1 \quad \text{and} \quad p = -1 \]
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**Summary:**
1. **Problem 1.3:** The equation \( 2(x - 3)^2 + 2 = 0 \) has **no real roots**; it has two **complex roots**.
2. **Problem 1.4:** The values of \( p \) that make the maximum value of \( g(x) = -2x^2 - px + 3 \) equal to \( 3\, \frac{1}{8} \) are \( p = 1 \) and \( p = -1 \).
Reviewed and approved by the UpStudy tutoring team
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