The equation \( 5 x^{2}+15 x+4=0 \) has two solutions \( A \) and \( B \) where \( A

Ask by Frazier Phillips. in the United States

Mar 09,2025

We start with the quadratic equation \[ 5x^2 + 15x + 4 = 0. \] **Step 1.** Identify the coefficients: \[ a = 5,\quad b = 15,\quad c = 4. \] **Step 2.** Compute the discriminant: \[ \Delta = b^2 - 4ac = 15^2 - 4 \cdot 5 \cdot 4 = 225 - 80 = 145. \] **Step 3.** Write down the quadratic formula: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-15 \pm \sqrt{145}}{2 \cdot 5} = \frac{-15 \pm \sqrt{145}}{10}. \] **Step 4.** Assign the solutions \(A\) and \(B\) such that \(A < B\). The smaller solution corresponds to the negative sign: \[ A = \frac{-15 - \sqrt{145}}{10}, \] and the larger solution corresponds to the positive sign: \[ B = \frac{-15 + \sqrt{145}}{10}. \] **Step 5.** For a decimal approximation (to 3 decimal places), note that \[ \sqrt{145} \approx 12.042. \] Then: \[ A \approx \frac{-15 - 12.042}{10} \approx \frac{-27.042}{10} \approx -2.704, \] \[ B \approx \frac{-15 + 12.042}{10} \approx \frac{-2.958}{10} \approx -0.296. \] The answers are: \[ A = \frac{-15 - \sqrt{145}}{10} \quad (\approx -2.704),\quad B = \frac{-15 + \sqrt{145}}{10} \quad (\approx -0.296). \]