SOLUTION: Find two integers whose product is 408 such that one of the integers is five less than seven times the other integer.

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Question 409556: Find two integers whose product is 408 such that one of the integers is five less than seven times the other integer.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

one of the integers is five less than seven times the other integer.

Let x and (7x-5)represent the intergers

Question states***

 x(7x-5) = 408

   7x^2 - 5x -408 = 0

Solving for x



x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%285+%2B-+sqrt%28+11449%29%29%2F%2814%29+

x+=+%285+%2B-+107%29%2F14

   x = 112/14 = 8  and x = -102/14 (extraneous solution as is a non-integer)  

Integers are 8 and 51 (7*8-5)



CHECKING our Answer***

   8*51 =408