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joshisabeast

  • 3 years ago

find all of the real or imaginary solutions of each equation by factoring 6x^3-2x^2+4x=0

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  1. Arcandin3
    • 3 years ago
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    6x^3-2x^2+4x=0 x(6x^2-2x+4) =0 now use the factoring formula (-b+- sqrt b^2 -4ac all over 2a) You get that the roots are x=0 x= (1/4)+2i(sqrt 23) x= (1/4)-2i(sqrt 23)

  2. joshisabeast
    • 3 years ago
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    thanks

  3. Arcandin3
    • 3 years ago
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    yup. Medal?

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