SOLUTION: find a number whose square is 36 greater than 5 times the number.

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Question 55654: find a number whose square is 36 greater than 5 times the number.
Answer by aaaaaaaa(138) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+=+5x+%2B+36
x%5E2+-+5x+-+36+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-36+=+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=%28-5%29%5E2-4%2A1%2A-36=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+169+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+169+%29%29%2F2%5C1+=+9
x%5B2%5D+=+%28-%28-5%29-sqrt%28+169+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B-5x%2B-36 can be factored:
1x%5E2%2B-5x%2B-36+=+1%28x-9%29%2A%28x--4%29
Again, the answer is: 9, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-36+%29