SOLUTION: An interger is four less than twice another. If the product of the two intergers is 70, one pair of intergers that will satisfy the problem is____.

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Question 950347: An interger is four less than twice another. If the product of the two intergers is 70, one pair of intergers that will satisfy the problem is____.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
F=first integer; S= second integer
FS=70
S=70/F
F-4=2S Substitute for S
F-4=2(70/F)
F-4=140/F Multiply each side by F.
F^2-4F=140 Subtract 140 from each side.
F^2-4F-140=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aF%5E2%2BbF%2Bc=0 (in our case 1F%5E2%2B-4F%2B-140+=+0) has the following solutons:

F%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-4%29%5E2-4%2A1%2A-140=576.

Discriminant d=576 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+576+%29%29%2F2%5Ca.

F%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+576+%29%29%2F2%5C1+=+14
F%5B2%5D+=+%28-%28-4%29-sqrt%28+576+%29%29%2F2%5C1+=+-10

Quadratic expression 1F%5E2%2B-4F%2B-140 can be factored:
1F%5E2%2B-4F%2B-140+=+1%28F-14%29%2A%28F--10%29
Again, the answer is: 14, -10. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B-140+%29

Answers F=14 and F=-10
S=70/F=70/14=5 ANSWER 1: One pair of integers that satisfies the problem is 14 and 5.
For F=-10
S=70/F=70/-10=-7 ANSWER 2: Another pair of numbers that satisfies the problem is -10 and -7.