SOLUTION: I am trying to help my daughter with the following and an stuck. I need answers and the steps to get there Find zeros of: 2x^3 -12x^2 - 6x

Algebra ->  Equations -> SOLUTION: I am trying to help my daughter with the following and an stuck. I need answers and the steps to get there Find zeros of: 2x^3 -12x^2 - 6x      Log On


   

Question 1153999: I am trying to help my daughter with the following and an stuck.
I need answers and the steps to get there
Find zeros of: 2x^3 -12x^2 - 6x
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find zeros of: 2x^3 -12x^2 - 6x
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2x^3 -12x^2 - 6x = 2x*(x^2 - 6x - 3)
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Either 2x = 0 or (x^2 - 6x - 3) = 0
2x = 0 --> x = 0
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x^2 - 6x - 3 cannot be factored. Use the quadratic equation or complete the square.
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(x^2 - 6x - 3) = 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-6x%2B-3+=+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=%28-6%29%5E2-4%2A1%2A-3=48.

Discriminant d=48 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+48+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+48+%29%29%2F2%5C1+=+6.46410161513775
x%5B2%5D+=+%28-%28-6%29-sqrt%28+48+%29%29%2F2%5C1+=+-0.464101615137754

Quadratic expression 1x%5E2%2B-6x%2B-3 can be factored:
1x%5E2%2B-6x%2B-3+=+%28x-6.46410161513775%29%2A%28x--0.464101615137754%29
Again, the answer is: 6.46410161513775, -0.464101615137754. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B-3+%29

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I can do completing the square if you like.
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x^2 - 6x - 3 = 0
x^2 - 6x = 3
x^2 - 6x + 9 = 3+9 = 12
(x-3)^2 = 12
x-3+=+sqrt%2812%29
x+=+3+%2B+sqrt%2812%29
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x-3+=+-sqrt%2812%29
x+=+3+-+sqrt%2812%29
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sqrt%2812%29+=+2sqrt%283%29
You can sub that if you like.