SOLUTION: Please help me solve: 1) x²+x-72 2) x²-15x+156 3) x²-6x-16 ,Thanks

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Question 253747: Please help me solve:
1) x²+x-72
2) x²-15x+156
3) x²-6x-16
,Thanks
Found 2 solutions by drk, richwmiller:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve. First, there are no =0 parts so we factor.
--
1) x²+x-72
(x+9)(x-8)
--
2) x²-15x+156
can't factor - prime.
--
3) x²-6x-16
(x-8)(x+2)

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x²-15x+156
neither 15 nor 256 are prime but because the discriminant is less than zero there are no real solutions
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-15x%2B156+=+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=%28-15%29%5E2-4%2A1%2A156=-399.

The discriminant -399 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 -399 is + or - sqrt%28+399%29+=+19.9749843554382.

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-15%2Ax%2B156+%29