+618 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?
+9466 Let's find the two points on the parabola that touch the line y = 5 .
y = x2 + 3x + 2 We want to find the x values when y is equal to 5 .
5 = x2 + 3x + 2 Subtract 5 from both sides of this equation.
0 = x2 + 3x - 3 Use the quadratic formula to find x .
x = \({-3 \pm \sqrt{21} \over 2}\)
So we know that the points \(({-3 + \sqrt{21} \over 2}\,,\,5)\) and \(({-3 - \sqrt{21} \over 2}\,,\,5)\) are the endpoints of one side of the square. Here's a graph that shows this: https://www.desmos.com/calculator/djyyxourv3
The side length of the square is the distance between these two points.
side length \(={-3 + \sqrt{21} \over 2} - {-3 - \sqrt{21} \over 2} \\~\\ ={-3 + \sqrt{21} \,-\,(-3-\sqrt{21})\over 2} \\~\\ ={-3 + \sqrt{21} + 3+\sqrt{21}\over 2} \\~\\ =\frac{2\sqrt{21}}{2} \\~\\ =\sqrt{21}\)
and...
area of square = (side length)2
area of square = \(\sqrt{21}^2\)
area of square = 21 square units
+9466 Let's find the two points on the parabola that touch the line y = 5 .
y = x2 + 3x + 2 We want to find the x values when y is equal to 5 .
5 = x2 + 3x + 2 Subtract 5 from both sides of this equation.
0 = x2 + 3x - 3 Use the quadratic formula to find x .
x = \({-3 \pm \sqrt{21} \over 2}\)
So we know that the points \(({-3 + \sqrt{21} \over 2}\,,\,5)\) and \(({-3 - \sqrt{21} \over 2}\,,\,5)\) are the endpoints of one side of the square. Here's a graph that shows this: https://www.desmos.com/calculator/djyyxourv3
The side length of the square is the distance between these two points.
side length \(={-3 + \sqrt{21} \over 2} - {-3 - \sqrt{21} \over 2} \\~\\ ={-3 + \sqrt{21} \,-\,(-3-\sqrt{21})\over 2} \\~\\ ={-3 + \sqrt{21} + 3+\sqrt{21}\over 2} \\~\\ =\frac{2\sqrt{21}}{2} \\~\\ =\sqrt{21}\)
and...
area of square = (side length)2
area of square = \(\sqrt{21}^2\)
area of square = 21 square units
