\( x+4=2 y \) and \( y^{2}-x y+21=0 \) 1.3 Discuss the nature of the roots of the equation \( 2(x-3)^{2}+2=0 \) \( 1.4 \quad \) Determine the value(s) of \( p \) if \( g(x)=-2 x^{2}-p x+3 \) has a maximum value of \( 3 \frac{1}{8} \)

Let's address each problem step by step. --- ### **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: \[ p = 1 \quad \text{and} \quad p = -1 \] --- **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 \( p = -1 \).