SOLUTION: the product of two consecutive integers is 25 less than 5 times their sum

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Question 948331: the product of two consecutive integers is 25 less than 5 times their sum
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
X=first integer; X+1=second integer
(X)(X+1)=5(X+(X+1))-25
X%5E2%2BX=5X%2B5X%2B5-25
X%5E2%2BX=10X-20 Subtract 10X from each side.
X%5E2-9X=-20 Add 20 to each side.
X%5E2-9X%2B20=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aX%5E2%2BbX%2Bc=0 (in our case 1X%5E2%2B-9X%2B20+=+0) has the following solutons:

X%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=%28-9%29%5E2-4%2A1%2A20=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--9%2B-sqrt%28+1+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%28-9%29%2Bsqrt%28+1+%29%29%2F2%5C1+=+5
X%5B2%5D+=+%28-%28-9%29-sqrt%28+1+%29%29%2F2%5C1+=+4

Quadratic expression 1X%5E2%2B-9X%2B20 can be factored:
1X%5E2%2B-9X%2B20+=+1%28X-5%29%2A%28X-4%29
Again, the answer is: 5, 4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-9%2Ax%2B20+%29

ANSWERS form X are 4 and 5
CHECK
For X=4, X+1=5
4(5)=5(4+5)-25
20=45-25
20=20 True, so 4 is a correct answer. ANSWER 1: One pair of consecutive integers is 4,5
For X=5, X+1=6
5(6)=5(5+6)-25
30=55-25
30=30 So 5 is also a correct answer. ANSWER 2: Another pair of consecutive integers is 5,6.
So there are two pairs of consecutive integers that fulfill these requirements, (4,5) and (5,6).