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)   (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

Subtract 10X from each side.
Add 20 to each side.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 5, 4. Here's your graph:

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).
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