SOLUTION: Find three consecutive even integers such that the product of the two smallest exceeds the largest by 38

Question 282532: Find three consecutive even integers such that the product of the two smallest exceeds the largest by 38
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find three consecutive even integers such that the product of the two smallest exceeds the largest by 38
.
Let x = 1st consecutive integer
then
x+2= 2nd consecutive integer
x+4 = 3rd consecutive integer
.
(x+1)(x+2) > (x+4)+38
x^2 + 3x + 2 > x + 42
x^2 + 2x + 2 > 42
x^2 + 2x - 40 > 0
Applying the quadratic equation we get:
x > 5.4 (next even integer is 6)
and
x > -7.4
.
Therefore, one set is:
6, 8 and 9
.
Details of quadratic follows:

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=164 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5.40312423743285, -7.40312423743285. Here's your graph:

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