SOLUTION: Please help me solve my problem {{{p^2-5p+81=p+4}}}

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Question 445088: Please help me solve my problem
p%5E2-5p%2B81=p%2B4
Answer by chriswen(106) About Me  (Show Source):
You can put this solution on YOUR website!
p%5E2-5p%2B81=p%2B4
p%5E2-5p%2B81-p-4=0
p%5E2-6p%2B77=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ap%5E2%2Bbp%2Bc=0 (in our case 1p%5E2%2B-6p%2B77+=+0) has the following solutons:

p%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=%28-6%29%5E2-4%2A1%2A77=-272.

The discriminant -272 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 -272 is + or - sqrt%28+272%29+=+16.4924225024706.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B77+%29

...
So basically this question is impossible or you could say it is a complex number if you have studied it.