SOLUTION: find the second of the three consecutive positive even integers such that the product of the first and second is 64 less than the square of the third

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Question 390059: find the second of the three consecutive positive even integers such that the product of the first and second is 64 less than the square of the third
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi,        

Let x, (x+2),(x+4) represent the three consecutive positive even integers

Question States***

 x(x+2) = (x+4)^2 - 64

solving for x

 x^2 + 2x = x^2 + 8x + 16 - 64

       48 = 6x

        x = 8, the first of the three.  The Second Integer is 10



CHECKING our Answer*****

 8*10 = 144 - 64 

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
 

given:



the three consecutive positive even integers 



Assign variables :



    Let  be the first consecutive positive even integers 



     be the  second consecutive positive even integers 



    be the third consecutive positive even integers 



the product of the first and second is  less than the square of the third















 







...or











.....................the first consecutive positive even integer





  



..............the  second consecutive positive even integer







...............the third consecutive positive even integer





check:

















 

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