+0  
 
0
34
5
avatar+184 

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. What is the area of the square? Please help, I'm terrible at quadratics!

 Jun 14, 2023
 #1
avatar
0

First, we need to find the two points on the parabola where y = 7. We can do this by setting y equal to 7 and solving for x. This gives us the following two equations:

2x^2 + 8x + 4 = 7 2x^2 + 8x - 3 = 0

We can solve this equation using the quadratic formula. This gives us the following two solutions:

x = -4 or x = -1/2

These are the two points where the side of the square intersects the parabola.

Now, we need to find the length of the side of the square. This is simply the distance between the two points we just found. This distance is equal to:

Code snippet

|(-4) - (-1/2)| = 9/2

Therefore, the area of the square is equal to:

(9/2)^2 = 81/4 = 20.25

So the answer is 20.25

 Jun 14, 2023
 #3
avatar+118609 
+3

You need to find out how long one side is.

 

The the ends of the side are the intersection of  y=7  and  y = 2x^2 + 8x + 4

 

Solve simultaneously

 

\(7=2x^2+8x+4\\ 2x^2+8x-3=0\\ x=\frac{-8\pm\sqrt{64+24}}{4}\\ x=\frac{-8\pm\sqrt{88}}{4}\\ x=\frac{-8\pm2\sqrt{22}}{4}\\ x=\frac{-8\pm2\sqrt{22}}{4}\\ x=-2\pm\frac{\sqrt{22}}{2}\\ \text{Subtract the little from the big and you get the side length of }\;\;\\\ \frac{2\sqrt{22}}{2}=\sqrt{22}\\ \text{So area is 22 }units^2\)

 Jun 14, 2023

4 Online Users

avatar
avatar
avatar