(t+2)²+(t-3)²=9 transform equations to quadratic equation in form ax²+bx+c=o

Answer:To transform the equation \((t+2)^2 + (t-3)^2 = 9\) into a quadratic equation of the form \(ax^2 + bx + c = 0\), follow these steps:1. **Expand each squared term**:   \[   (t+2)^2 = t^2 + 4t + 4   \]   \[   (t-3)^2 = t^2 - 6t + 9   \]2. **Add the expanded terms**:   \[   (t+2)^2 + (t-3)^2 = (t^2 + 4t + 4) + (t^2 - 6t + 9)   \]   Combine like terms:   \[   t^2 + 4t + 4 + t^2 - 6t + 9 = 2t^2 - 2t + 13   \]3. **Set the equation equal to 9**:   \[   2t^2 - 2t + 13 = 9   \]4. **Subtract 9 from both sides to set the equation to 0**:   \[   2t^2 - 2t + 13 - 9 = 0   \]   \[   2t^2 - 2t + 4 = 0   \]So, the quadratic equation in the form \(ax^2 + bx + c = 0\) is:\[2t^2 - 2t + 4 = 0\]Step-by-step explanation:

Answered by jumongverano | 2024-09-03