SOLUTION: Find three consecutive positive integers such that twice the product of the two smaller integers is 88 more than the product of the two larger integers.

Question 316961: Find three consecutive positive integers such that twice the product of the two smaller integers is 88 more than the product of the two larger integers.
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
x,x+2,x+4
x and x+2 are the 2 smaller numbers
"twice the product of the two smaller integers is 88 more than the product of the two larger integers"
add
subtract

doesnt come out to a whole number so either thats suppose to be that way in the answer or the question was written wrong.
x= 10.8488578017961 or -8.8488578017961
(x+2)= 12.8488578017961 or -6.8488578017961
(x+4)= 14.8488578017961 or -4.8488578017961

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=388 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 10.8488578017961, -8.8488578017961. Here's your graph:

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