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)   (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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

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 - .

The solution is

Here's your graph:
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