The equation \( 2 x^{2}+15 x+1=0 \) has two solutions \( A \) and \( B \) where \( A

Ask by Ortega Osborne. in the United States

Feb 27,2025

To find the solutions \( A \) and \( B \) for the equation \( 2x^{2} + 15x + 1 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, \( a = 2 \), \( b = 15 \), and \( c = 1 \). Plugging these values into the formula: 1. Calculate the discriminant: \[ b^{2} - 4ac = 15^{2} - 4 \cdot 2 \cdot 1 = 225 - 8 = 217 \] 2. Now we substitute back into the quadratic formula: \[ x = \frac{-15 \pm \sqrt{217}}{2 \cdot 2} = \frac{-15 \pm \sqrt{217}}{4} \] From this, we have the two solutions: \[ A = \frac{-15 - \sqrt{217}}{4} \quad \text{and} \quad B = \frac{-15 + \sqrt{217}}{4} \] Now, if we want to provide these as decimal approximations: Calculating \( \sqrt{217} \approx 14. rejoinder.931 \) gives us: \[ A \approx \frac{-15 - 14.931}{4} \approx \frac{-29.931}{4} \approx -7.483 \] \[ B \approx \frac{-15 + 14.931}{4} \approx \frac{-0.069}{4} \approx -0.017 \] Thus, the solutions are: \( A \approx -7.483 \) and \( B \approx -0.017 \) So, rounding to three decimal places: \( A = -7.483 \) and \( B = -0.017 \)