SOLUTION: Find 2 positive real numbers that differ by 2 and have a product of 10.

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Question 298679: Find 2 positive real numbers that differ by 2 and have a product of 10.
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Find 2 positive real numbers that differ by 2 and have a product of 10.
.
Let x = smaller of two positive real number
then
x+2 = larger of two positive real number
.
x(x+2) = 10
x^2+2x = 10
x^2+2x-10 = 0
.
Since we can't factor, we must resort to the quadratic formula giving us:
x = {2.317, -4.317}
Since the problem want "positive" numbers our solution is:
x = 2.317
The other number is then
2.317 + 2 = 4.317
.
The two numbers are 2.317 and 4.317
.
Details of quadratic formula follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-10+=+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-10=44.

Discriminant d=44 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+44+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+44+%29%29%2F2%5C1+=+2.3166247903554
x%5B2%5D+=+%28-%282%29-sqrt%28+44+%29%29%2F2%5C1+=+-4.3166247903554

Quadratic expression 1x%5E2%2B2x%2B-10 can be factored:
1x%5E2%2B2x%2B-10+=+1%28x-2.3166247903554%29%2A%28x--4.3166247903554%29
Again, the answer is: 2.3166247903554, -4.3166247903554. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-10+%29