### Instruction: A square is drawn such that one of its sides coincides with the line $y = 5$, and so that the endpoints of this side lie on the parabola $y = x^2 + 3x + 2$. What is the area of the square?### Response: To find the area of the square, I need to know the length of its side.
The side of the square that coincides with the line $y = 5$ must have the same $y$-coordinate for both endpoints.
So I can use the equation $y = x^2 + 3x + 2$ to find the $x$-coordinates of those endpoints.
I set $y = 5$ and get $5 = x^2 + 3x + 2$, which simplifies to $x^2 + 3x - 3 = 0$.
This is a quadratic equation that I can solve by using the quadratic formula.
I get $x = \frac{-3 \pm \sqrt{9 + 12}}{2} = \frac{-3 \pm \sqrt{21}}{2}$.
These are the $x$-coordinates of the endpoints of the side of the square on the parabola.
To find the length of the side, I need to subtract the smaller $x$-coordinate from the larger one.
This gives me $s = \frac{-3 + \sqrt{21}}{2} - \frac{-3 - \sqrt{21}}{2} = \sqrt{21}$.
Therefore, the area of the square is $s^2 = (\sqrt{21})^2 = 21$.