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FreeTrader

  • 4 years ago

Can you help me see where I am stuck finding the zeros of this quadratic equation?

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  1. FreeTrader
    • 4 years ago
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    \[3x^2 - 12x + 11 = 0\]

  2. anonymous
    • 4 years ago
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    quadratic formula for this one use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=3,b=-12,c=11\]

  3. FreeTrader
    • 4 years ago
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    My student solutions manual says I should obtain \[(6 \pm \sqrt{3}) \div 3\]

  4. anonymous
    • 4 years ago
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    indeed you should

  5. FreeTrader
    • 4 years ago
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    Let's walk through this, so I can see where I am tripping over my shoelaces.

  6. anonymous
    • 4 years ago
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    first you get \[\frac{-12\pm\sqrt{(-12)^2-4\times 3\times 11}}{6}\]

  7. FreeTrader
    • 4 years ago
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    Wait a sec. I see that I made an error in the first term under the radical. I will go back to the drawing board and re-attempt this one on my own.

  8. FreeTrader
    • 4 years ago
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    By the way, how are you producing the nice looking divisor line?

  9. anonymous
    • 4 years ago
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    \frac{a}{b} looks like \[\frac{a}{b}\]

  10. FreeTrader
    • 4 years ago
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    Got it. I made simple mistakes in my substitutions, then got hung up on trying to do algebra with unduly complicated figures.

  11. FreeTrader
    • 4 years ago
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    THANKS @satellite73, you got me unstuck!

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