SOLUTION: Find two integers which differ by 8 such that their product is-15.

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Question 697467: Find two integers which differ by 8 such that their product is-15.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
two integers differ by 8: x-y = 8

solve for y: y = x-8

their product is -15: xy = -15

plug in y = x-8: x(x-8) = -15

Now solve for x

x(x-8) = -15

x^2 - 8x = -15

x^2 - 8x + 15 = 0

Use the quadratic formula to solve for x

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

x+=+%28-%28-8%29%2B-sqrt%28%28-8%29%5E2-4%281%29%2815%29%29%29%2F%282%281%29%29 Plug in a+=+1, b+=+-8, c+=+15

x+=+%288%2B-sqrt%2864-%2860%29%29%29%2F%282%29

x+=+%288%2B-sqrt%284%29%29%2F2

x+=+%288%2Bsqrt%284%29%29%2F2 or x+=+%288-sqrt%284%29%29%2F2

x+=+%288%2B2%29%2F2 or x+=+%288-2%29%2F2

x+=+10%2F2 or x+=+6%2F2

x+=+5 or x+=+3

Now that you know x = 5, you can use this to find that y = -3

Or if x = 3, then you can find out that y = -5. Both of these y values are found when you plug in the corresponding x value into y+=+x+-+8 and you solve for y.

So the two numbers are 5 and -3. Also, another pair of numbers is 3 and -5