SOLUTION: Find two integers whose product is 105 such that one of the integers is one more than twice the other integer.

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Question 965436: Find two integers whose product is 105 such that one of the integers is one more than twice the other integer.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
x and y are the integers
x=2y+1
xy=105 Substitute for x
(2y+1)(y)=105 Subtract 105 from each side
2y^2+y-105=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 2y%5E2%2B1y%2B-105+=+0) has the following solutons:

y%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=%281%29%5E2-4%2A2%2A-105=841.

Discriminant d=841 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+841+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%281%29%2Bsqrt%28+841+%29%29%2F2%5C2+=+7
y%5B2%5D+=+%28-%281%29-sqrt%28+841+%29%29%2F2%5C2+=+-7.5

Quadratic expression 2y%5E2%2B1y%2B-105 can be factored:
2y%5E2%2B1y%2B-105+=+2%28y-7%29%2A%28y--7.5%29
Again, the answer is: 7, -7.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-105+%29

y=7 ANSWER 1: One of the integers is 7
x=2y+1=2(7)+1=14+1=15 ANSWER 2: The other integer is 15