hectictar
Apr 15, 2018

#1**+1 **

Let's find the two points on the parabola that touch the line y = 5 .

y = x^{2} + 3x + 2 We want to find the x values when y is equal to 5 .

5 = x^{2} + 3x + 2 Subtract 5 from both sides of this equation.

0 = x^{2} + 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

hectictar
Jul 17, 2017