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# help plz!

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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?

Aug 25, 2023

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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?

Aug 25, 2023

#1
+177
+1

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?

The3Mathketeers Aug 25, 2023