Question 1119464: find two numbers whose difference is 3 and whose product is 88
Found 2 solutions by math_helper, Alan3354:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! Let x = the larger number
Then (x)(x-3) = 88
x=11 and/or x=-8
x=11 —> x-3=8, 11*8 = 88…  is a solution
x=-8 —> x-3 = -11, -8*-11=88 …  is a solution
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Tutor @Alan3354 must have misspoken, as a quadratic equation most certainly may have integer solutions.
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No.
I said if the 2 solutions are NOT integers, then solving a quadratic is necessary.
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If they are integers, forming a quadratic and then factoring it just adds unneeded lines IF the solutions are integers.
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PS I intended to edit my answer, but for some reason I'm editing someone else's ???
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To Alan5534 - Yeah, weird, it allowed you to edit my post. Anyway, I get what you are saying now. However, I do think forming the quadratic and then solving is a straightforward method that works every time (at least when there is a real answer), while the guessing at factors may or may not work, depending on the solution. Thanks.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! find two numbers whose difference is 3 and whose product is 88
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Find a pair of factors of 88 that differ by 3.
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1*88 NG
2*44 NG
etc.
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If the numbers are not integers, you would have do solve a quadratic equation.
These 2 are integers, so making a quadratic,
x*(x-3) = 88 --> x^2 - 3x - 88 = 0 does not help.
At this point, to factor, you find a pair of factors of 88 that differ by 3.
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