You're not done with the first question, you were to show that 2b^2=9c

Question

You're not done with the first question, you were to show that 2b^2=9c

Asked on April 24, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-24T19:42:28.421Z

Step 1: Apply Vieta's formulas for the given quadratic equation. For the quadratic equation $2x^2 + bx + c = 0$, the sum of the roots ($\alpha + \beta$) and the product of the roots ($\alpha \beta$) are given by: $$ \alpha + \beta = -\frac{b}{2} \quad (1) $$ $$ \alpha \beta = \frac{c}{2} \quad (2) $$ Step 2: Use the given condition $\alpha = 2\beta$ in the sum of roots equation. Substitute $\alpha = 2\beta$ into equation (1): $$ 2\beta + \beta = -\frac{b}{2} $$ $$ 3

Accepted Answer · 2026-04-24T19:42:28.421Z

Step 1: Apply Vieta's formulas for the given quadratic equation. For the quadratic equation $2x^2 + bx + c = 0$, the sum of the roots ($\alpha + \beta$) and the product of the roots ($\alpha \beta$) are given by: $$ \alpha + \beta = -\frac{b}{2} \quad (1) $$ $$ \alpha \beta = \frac{c}{2} \quad (2) $$ Step 2: Use the given condition $\alpha = 2\beta$ in the sum of roots equation. Substitute $\alpha = 2\beta$ into equation (1): $$ 2\beta + \beta = -\frac{b}{2} $$ $$ 3

You're not done with the first question, you were to show that 2b^2=9c