SOLUTION: (x-3)²+(x+2)²+17

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Question 174386: (x-3)²+(x+2)²+17
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-3%29%5E2%2B%28x%2B2%29%5E2%2B17=0 Should be equal to "0" to make it an eqn.
Let's expand:
x%5E2-6x%2B9%2Bx%5E2%2B4x%2B4%2B17=0
2x%5E2-2x%2B30=0, divide whole eqn by 2:
x%5E2-x%2B15=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-1x%2B15+=+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=%28-1%29%5E2-4%2A1%2A15=-59.

The discriminant -59 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -59 is + or - sqrt%28+59%29+=+7.68114574786861.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B15+%29


Thank you,
Jojo