+721 A square is drawn such that one of its sides coincides with the line $y = 7$, and so that the endpoints of this side lie on the parabola $y = 2x^2 + 8x + 4 - 3x + 3$. What is the area of the square?
When x = 0, y = 7. How can I use that?
+177 One of the equations is y = 7 and the other equation is y = 2x2 + 8x + 4 - 3x + 3. Since both equations are solved for y, you can set each equation equal to each other and solve for the corresponding x-variable.
\(y = 7; y = 2x^2 + 8x + 4 - 3x + 3\\ 2x^2 + 8x + 4 - 3x + 3 = 7 \\ 2x^2 + 5x + 7 = 7\\ 2x^2 + 5x = 0 \\ x(2x + 5) = 0 \\ x = 0 \text{ or } 2x + 5 = 0 \\ x = 0 \text{ or } x = -\frac{5}{2}\)
As you mentioned, when x = 0, y = 7 as we found. This means that the coordinates of the endpoints are \((0, 7) \text{ and } \left(-\frac{5}{2}, 7\right)\). Do you know how to carry on from here?
+177 One of the equations is y = 7 and the other equation is y = 2x2 + 8x + 4 - 3x + 3. Since both equations are solved for y, you can set each equation equal to each other and solve for the corresponding x-variable.
\(y = 7; y = 2x^2 + 8x + 4 - 3x + 3\\ 2x^2 + 8x + 4 - 3x + 3 = 7 \\ 2x^2 + 5x + 7 = 7\\ 2x^2 + 5x = 0 \\ x(2x + 5) = 0 \\ x = 0 \text{ or } 2x + 5 = 0 \\ x = 0 \text{ or } x = -\frac{5}{2}\)
As you mentioned, when x = 0, y = 7 as we found. This means that the coordinates of the endpoints are \((0, 7) \text{ and } \left(-\frac{5}{2}, 7\right)\). Do you know how to carry on from here?
