The equation \( x^{2}+9 x-3=0 \) has two solutions \( A \) and \( B \) where \( A

Ask by Bird Stephens. in the United States

Jan 23,2025

To solve the equation \( x^{2}+9x-3=0 \), we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 9 \), and \( c = -3 \). First, we calculate the discriminant: \[ b^2 - 4ac = 9^2 - 4(1)(-3) = 81 + 12 = 93. \] Now plug the values into the formula: \[ x = \frac{-9 \pm \sqrt{93}}{2(1)} = \frac{-9 \pm \sqrt{93}}{2}. \] Calculating the two solutions: 1. \( A = \frac{-9 - \sqrt{93}}{2} \) 2. \( B = \frac{-9 + \sqrt{93}}{2} \) Now we can give the decimal values of \( A \) and \( B \): Using \( \sqrt{93} \approx 9.644 \): 1. \( A \approx \frac{-9 - 9.644}{2} \approx \frac{-18.644}{2} \approx -9.322 \) 2. \( B \approx \frac{-9 + 9.644}{2} \approx \frac{0.644}{2} \approx 0.322 \) Thus, the solutions are approximately: \( A \approx -9.322 \) and \( B \approx 0.322 \). The answers are: \( A = -9.322 \) and \( B = 0.322 \) (to three decimal places).