SOLUTION: If one end of a line whose length is 13 Units is the point (4, 8) and the ordinate of the other end is 3. What is its abscissa?

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Question 861827: If one end of a line whose length is 13 Units is the point (4, 8)
and the ordinate of the other end is 3. What is its abscissa?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If one end of a line whose length is 13 Units is the point (4, 8)
and the ordinate of the other end is 3. What is its abscissa?
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Make (4,8) the center of a circle of radius 13
--> %28x-4%29%5E2+%2B+%28y-8%29%5E2+=+169
Find the intersection(s) of x=3 and the circle.
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%28x-4%29%5E2+%2B+%28y-8%29%5E2+=+169
%283-4%29%5E2+%2B+%28y-8%29%5E2+=+169
y%5E2+-+16y+%2B+65+=+169
y%5E2+-+16y+-+104+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-16x%2B-104+=+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=%28-16%29%5E2-4%2A1%2A-104=672.

Discriminant d=672 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--16%2B-sqrt%28+672+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-16%29%2Bsqrt%28+672+%29%29%2F2%5C1+=+20.9614813968157
x%5B2%5D+=+%28-%28-16%29-sqrt%28+672+%29%29%2F2%5C1+=+-4.96148139681572

Quadratic expression 1x%5E2%2B-16x%2B-104 can be factored:
1x%5E2%2B-16x%2B-104+=+%28x-20.9614813968157%29%2A%28x--4.96148139681572%29
Again, the answer is: 20.9614813968157, -4.96148139681572. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-16%2Ax%2B-104+%29

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y+=+8+%2B-+2sqrt%2842%29