SOLUTION: A number differs from it's square by the average of the number and it's square

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Question 22673: A number differs from it's square by the average of the number and it's square
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number.
x%5E2 = square of the number
x%5E2+-+x = difference between a number and its square
%28x%2Bx%5E2%29%2F2+ = average of a number and it's square (add together and divide by 2)

The equation should look like this:
x%5E2+-+x+=+%28x%2Bx%5E2%29%2F2

Multiply both sides of the equation by 2:
2%2A%28x%5E2+-+x%29+=+2%2A%28%28x%2Bx%5E2%29%2F2%29
2x%5E2+-+2x+=+x+%2B+x%5E2

Set equal to zero, by subtracting x and x%5E2 from each side:
2x%5E2+-+x%5E2+-+2x+-+x+=+x%2Bx%5E2+-x%5E2+-+x+=0
x%5E2+-+3x+=+0

Factor out the common factor of x:
x%28x-3%29+=+0
x= 0 or x= 3

There are two solutions.
x=0: The square is 0. The difference of the number and its square is zero, and the average of 0 and its square is also 0.

x= 3: The square is 9. The difference of the number and its square is 6, and the average of the number and its square is %283%2B9%29%2F2=+6.

Both answers check.

R^2 at SCC