SOLUTION: give exact and approximate solution to three decimal places x^2+14x+49=1

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Question 545590: give exact and approximate solution to three decimal places
x^2+14x+49=1
Answer by jerryguo41(197) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B14x%2B49=1
1) Move one over by subtracting 1 on both sides
{{x^2+14x+48=0}}}
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B14x%2B48+=+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=%2814%29%5E2-4%2A1%2A48=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-14%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2814%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+-6
x%5B2%5D+=+%28-%2814%29-sqrt%28+4+%29%29%2F2%5C1+=+-8

Quadratic expression 1x%5E2%2B14x%2B48 can be factored:
1x%5E2%2B14x%2B48+=+1%28x--6%29%2A%28x--8%29
Again, the answer is: -6, -8. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B14%2Ax%2B48+%29