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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[3x^2  12x + 11 = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0quadratic formula for this one use \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\] with \[a=3,b=12,c=11\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My student solutions manual says I should obtain \[(6 \pm \sqrt{3}) \div 3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's walk through this, so I can see where I am tripping over my shoelaces.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first you get \[\frac{12\pm\sqrt{(12)^24\times 3\times 11}}{6}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait 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 reattempt this one on my own.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By the way, how are you producing the nice looking divisor line?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\frac{a}{b} looks like \[\frac{a}{b}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Got it. I made simple mistakes in my substitutions, then got hung up on trying to do algebra with unduly complicated figures.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0THANKS @satellite73, you got me unstuck!
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