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Question 1026033: Find two consecutive odd integers such that their product is 47 more than 4 times their sum.
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! integers are x and x+2
x(x+2)=4(2x+2)+47
x^2+2x=8x+8+47
x^2+2x=8x+55
x^2-6x-55=0
(x-11)(x+5)=0
x=11,-5
11 and 13 has a sum of 24 and a product of 143. That is 4*24+47
-5 and -3 has a sum of -8 and a product of 15. That is 4*(-8)+47
The pairs are 11,13 and -5,-3.
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