SOLUTION: "what are the real or imaginary solutions of 4x3-6x2=4x" x(2x-3x)(2x-2x)

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Question 598950: "what are the real or imaginary solutions of 4x3-6x2=4x"
x(2x-3x)(2x-2x)

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
"what are the real or imaginary solutions of 4x3-6x2=4x"
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4x3-6x2-4x = 0
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2x%2A%282x%5E2+-+3x+-+2%29+=+0
x = 0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-3x%2B-2+=+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=%28-3%29%5E2-4%2A2%2A-2=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+25+%29%29%2F2%5C2+=+2
x%5B2%5D+=+%28-%28-3%29-sqrt%28+25+%29%29%2F2%5C2+=+-0.5

Quadratic expression 2x%5E2%2B-3x%2B-2 can be factored:
2x%5E2%2B-3x%2B-2+=+%28x-2%29%2A%28x--0.5%29
Again, the answer is: 2, -0.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-3%2Ax%2B-2+%29