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FreeTrader
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Can you help me see where I am stuck finding the zeros of this quadratic equation?
 3 years ago
 3 years ago
FreeTrader Group Title
Can you help me see where I am stuck finding the zeros of this quadratic equation?
 3 years ago
 3 years ago

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FreeTrader Group TitleBest ResponseYou've already chosen the best response.0
\[3x^2  12x + 11 = 0\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
quadratic formula for this one use \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\] with \[a=3,b=12,c=11\]
 3 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
indeed you should
 3 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
first you get \[\frac{12\pm\sqrt{(12)^24\times 3\times 11}}{6}\]
 3 years ago

FreeTrader Group TitleBest ResponseYou've already chosen the best response.0
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 reattempt this one on my own.
 3 years ago

FreeTrader Group TitleBest ResponseYou've already chosen the best response.0
By the way, how are you producing the nice looking divisor line?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\frac{a}{b} looks like \[\frac{a}{b}\]
 3 years ago

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

FreeTrader Group TitleBest ResponseYou've already chosen the best response.0
THANKS @satellite73, you got me unstuck!
 3 years ago
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