1.4 The solutions of a quadratic equation are given by \( x=\frac{-2 \pm \sqrt{2 p+5}}{7} \) State the value(s) of \( p \) for which this equation will have: \( \begin{array}{ll}\text { 1.4.1 } & \text { Two equal solutions } \\ \text { 1.4.2 } & \text { No real solutions }\end{array} \)

To determine the values of \( p \) for which the given quadratic equation has two equal solutions or no real solutions, let's analyze the equation: \[ x = \frac{-2 \pm \sqrt{2p + 5}}{7} \] ### Understanding the Quadratic Equation A general quadratic equation is given by: \[ ax^2 + bx + c = 0 \] The solutions (roots) of this equation are: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Comparing this with the given equation: \[ x = \frac{-2 \pm \sqrt{2p + 5}}{7} \] We can identify: - \( 2a = 7 \) ⇒ \( a = \frac{7}{2} \) - \( -b = -2 \) ⇒ \( b = 2 \) - The discriminant \( D = 2p + 5 \) corresponds to \( b^2 - 4ac \). ### 1.4.1 Two Equal Solutions Two equal (repeated) solutions occur when the discriminant is zero: \[ D = 2p + 5 = 0 \] Solving for \( p \): \[ 2p + 5 = 0 \\ 2p = -5 \\ p = -\frac{5}{2} \] **Thus, the equation has two equal solutions when \( p = -\frac{5}{2} \).** ### 1.4.2 No Real Solutions No real solutions occur when the discriminant is negative: \[ D = 2p + 5 < 0 \] Solving for \( p \): \[ 2p + 5 < 0 \\ 2p < -5 \\ p < -\frac{5}{2} \] **Thus, the equation has no real solutions when \( p < -\frac{5}{2} \).** ### Summary 1. **Two equal solutions:** \( p = -\frac{5}{2} \) 2. **No real solutions:** \( p < -\frac{5}{2} \) Answer: 1.4.1 Two equal solutions when p = –5⁄2; 1.4.2 No real solutions for all real numbers p with p < –5⁄2.