Find two consecutive odd integers such that their product is 19 more than 4 times their sum
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Represent the smaller odd number as 2x-1 where x is a whole number. The second number will be 2x+1 because 2x-1+2=2x+1. The product is (2x-1)(2x+1)=4x²-1. Their sum is 2x-1+2x+1=4x. 4 times their sum is 16x.

Therefore 4x²-1=16x+19, that is, 4x²-16x-20=0 and we can divide through by 4:

x²-4x-5=0=(x+1)(x-5) so x is -1 or 5.

We can now work out the odd numbers. 2x-1=-3, so the other number is 2 more than this, making it -1.

Or when we have x=5, the numbers are 10-1=9 and 11.

Let’s see if they’re right. -3 times -1=3, and -3 plus -1 is -4; 4 times this is -16, and 19 more makes 3, which is their product so -1 and -3 is a solution.

9 times 11=99, 9 plus 11=20, and 4 times 20=80. So 99=80+19 is true, making 9 and 11 a solution.

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