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AshhSmith

a^2 – 5a – 20 factor completely

  • one year ago
  • one year ago

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  1. AshhSmith
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    -5(4) right ?

    • one year ago
  2. vishweshshrimali5
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    No not this time I think you have copied your question incorrectly it should be a^2 - a - 20 right ?

    • one year ago
  3. AshhSmith
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    no it has 5 in this one

    • one year ago
  4. vishweshshrimali5
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    ok

    • one year ago
  5. AshhSmith
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    i see it's prime so how do i show work

    • one year ago
  6. nitz
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    is the answer \[5+\sqrt{105}/2,5-\sqrt{105}/2\]

    • one year ago
  7. precal
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    when it is prime, we just state "prime" and do not attempt to factor. Now if it said, solve then we solve using the quadratic formula

    • one year ago
  8. nitz
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    7.623,-2.623

    • one year ago
  9. nitz
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    yes or no?

    • one year ago
  10. AshhSmith
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    no because it's prime

    • one year ago
  11. nitz
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    but it can be solved by discriminant method

    • one year ago
  12. AshhSmith
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    it just said factor and show work so i don't think i have to use that .

    • one year ago
  13. precal
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    correct, if the instructions were to factor write "prime" if the instructions were to solve then use the quadratic formula

    • one year ago
  14. nitz
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    It's prime! Applying quadratic formula, we get: a^2 – 5*a – 20 = 0 a = 1, b = - 5, c = - 20 b^2 - 4*a*c = 25 - 4*1*( - 20) = 105 a1 = ( - ( - 5) + sqrt(105))/2 = 5/2 + (1/2)*sqrt(105) a2 = 5/2 - (1/2)*sqrt(105)

    • one year ago
  15. nitz
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    ya??got it!

    • one year ago
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