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anonymous

  • one year ago

URGENT HElP Where did I go wrong in my problem?!?!?

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  1. anonymous
    • one year ago
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  2. UsukiDoll
    • one year ago
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    so we have the quadratic equation \[x^2+6x+3\] using the discriminant formula \[ b^2-4ac \] we have \[(6)^2-4(1)(3) =36-12 =24\] which is not a perfect square. I need to grab the latex for the quadratic equation brb

  3. anonymous
    • one year ago
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    ok, I'll be here

  4. UsukiDoll
    • one year ago
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    \[{x = \frac{{ - 6 \pm \sqrt {24} }}{{2}}} \]

  5. UsukiDoll
    • one year ago
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    ok the problem lies in the square root... yes it is 24, but we can simplify that further.

  6. UsukiDoll
    • one year ago
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    so we need to find the smallest perfect square number

  7. anonymous
    • one year ago
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    Oh I get it, so we can do √6 √4 which simplifies into 2√6 which would go away when divided by 2. right?

  8. UsukiDoll
    • one year ago
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    yeah

  9. anonymous
    • one year ago
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    Also how do I explain why I chose this method?

  10. UsukiDoll
    • one year ago
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    \[{x = \frac{{ - 6 \pm \sqrt {4 \cdot 6 } }}{{2}}} \rightarrow {x = \frac{{ - 6 \pm 2\sqrt {6} }}{{2}}}\] \[x = \frac{-6}{2} \pm \frac{2\sqrt{6}}{2} \rightarrow -3 \pm \sqrt{6}\]

  11. UsukiDoll
    • one year ago
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    First of all, we can't factor this equation. Using the discriminant formula, we obtain 24 and that's not a perfect square number. Therefore, we have to use the quadratic formula.

  12. UsukiDoll
    • one year ago
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    Completing the square is useless because we are given a quadratic equation which is similar to the standard quadratic formula of \[ax^2+bx+c \]

  13. UsukiDoll
    • one year ago
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    I don't think we can do graphing either... our equation doesn't start with y =... that's kind of a generic excuse though.

  14. anonymous
    • one year ago
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    Ok thank you SO much

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spraguer (Moderator)
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