SOLUTION: for the quadraic equation x^2-5x+5=0 give the exact and approximate solution to three decimal places.

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Question 317953: for the quadraic equation x^2-5x+5=0 give the exact and approximate solution to three decimal places.
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B5+=+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%2A5=5.

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

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+5+%29%29%2F2%5C1+=+3.61803398874989
x%5B2%5D+=+%28-%28-5%29-sqrt%28+5+%29%29%2F2%5C1+=+1.38196601125011

Quadratic expression 1x%5E2%2B-5x%2B5 can be factored:
1x%5E2%2B-5x%2B5+=+1%28x-3.61803398874989%29%2A%28x-1.38196601125011%29
Again, the answer is: 3.61803398874989, 1.38196601125011. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B5+%29