Solve for \( p \frac{5 p_{3}^{-2}}{3}+\frac{p}{2}=8 \)

To solve the equation \( p \frac{5 p_{3}^{-2}}{3}+\frac{p}{2}=8 \), we first want to clarify what \( p_3^{-2} \) means. If we take \( p_3^{-2} \) to mean \( p^{-2} \), our equation becomes: \[ p \frac{5 p^{-2}}{3} + \frac{p}{2} = 8 \] Now we'll rewrite the left side: \[ \frac{5p \cdot p^{-2}}{3} + \frac{p}{2} = 8 \] This simplifies to: \[ \frac{5}{3p} + \frac{p}{2} = 8 \] Next, we want to eliminate the fractions. To do this, we can multiply through the equation by \( 6p \) (the least common multiple of the denominators): \[ 6p \left(\frac{5}{3p}\right) + 6p \left(\frac{p}{2}\right) = 6p \cdot 8 \] This simplifies to: \[ 2 \cdot 5 + 3p^2 = 48p \] or \[ 10 + 3p^2 = 48p \] Next, we rearrange the equation: \[ 3p^2 - 48p + 10 = 0 \] Now, we can use the quadratic formula, \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -48 \), and \( c = 10 \): Calculate the discriminant: \[ b^2 - 4ac = (-48)^2 - 4 \cdot 3 \cdot 10 = 2304 - 120 = 2184 \] Now, substitute back into the quadratic formula: \[ p = \frac{-(-48) \pm \sqrt{2184}}{2 \cdot 3} = \frac{48 \pm \sqrt{2184}}{6} \] Now calculate \( \sqrt{2184} \): \[ \sqrt{2184} \approx 46.707 (exact values can be calculated too) \] Thus, we find: \[ p = \frac{48 \pm 46.707}{6} \] Calculating the two possible values: 1. \( p = \frac{48 + 46.707}{6} \approx 15.1178 \) 2. \( p = \frac{48 - 46.707}{6} \approx 0.2155 \) Therefore, the solutions for \( p \) are approximately \( 15.12 \) and \( 0.22 \).