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

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Question 282532: Find three consecutive even integers such that the product of the two smallest exceeds the largest by 38
Answer by nerdybill(7384) About Me  (Show Source):
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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 ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-40+=+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=%282%29%5E2-4%2A1%2A-40=164.

Discriminant d=164 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+164+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+164+%29%29%2F2%5C1+=+5.40312423743285
x%5B2%5D+=+%28-%282%29-sqrt%28+164+%29%29%2F2%5C1+=+-7.40312423743285

Quadratic expression 1x%5E2%2B2x%2B-40 can be factored:
1x%5E2%2B2x%2B-40+=+1%28x-5.40312423743285%29%2A%28x--7.40312423743285%29
Again, the answer is: 5.40312423743285, -7.40312423743285. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-40+%29