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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 \(y = 2x^2 + 8x + 4\). What is the area of the square?

 Jul 16, 2019
 #1
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We just need to find the x intersection points  of the parabola and the line y  = 7

 

So....we have that

 

7  = 2x^2 + 8x + 4       subtract 7 from both sides

 

2x^2 + 8x - 3  = 0     

 

Uing the quadratic formula

 

x  =  -8 ±√ [ 8^2 - 4(2)(-3) ]       =       -8 ±√ [ 88 ]      =    -8 ± 2√ 22    =     -2 ±√22 /2  =  - 4 ±√22

       __________________             __________         _________                                  ________

                 2 * 2                                      4                          4                                                 2

 

So....the two points  of  intersection   are  - 4 + √22      and  - 4 - √22

                                                                  _______             _______

                                                                        2                         2

 

And the square of the distance between these points  =  the area of the square  =

 

[  (  - 4 + √22) / 2   -  ( -4 - √22) / 2 ] ^2   =

 

[   2 √22 / 2 ] ^2  =

 

[ √22] ^2  =

 

22 units^2

 

 

cool cool cool

 Jul 16, 2019

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