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kitsune0724

  • 2 years ago

3p^2+12p-2=0 how do you solve this problem? I need all the steps please and thank you.

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  1. agentx5
    • 2 years ago
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    You have to use the Quadradic Formula here, it's not solvable by factoring: \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

  2. agentx5
    • 2 years ago
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    To be clear that's for the standard form: ax^2 + bx + c = 0 a = 3 b = 12 c = -2 See how that works? Yes, you will get two answers, as you should (because of the x\(^2\))

  3. kitsune0724
    • 2 years ago
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    okay

  4. agentx5
    • 2 years ago
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    If what is under the radical/square-root above is negative you'd have two imaginary roots, thankfully you won't here. But you will have square-roots (or fractional powers if you prefer) in your final answers.

  5. kitsune0724
    • 2 years ago
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    okay

  6. kitsune0724
    • 2 years ago
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    is it \[x=12\pm \sqrt{120} \over 6\]

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