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anonymous
 4 years ago
I can't figure out what I'm doing wrong. Solve the equation by completing the square to obtain the exact solution. 3xsquare + 6x + 2 = 0
anonymous
 4 years ago
I can't figure out what I'm doing wrong. Solve the equation by completing the square to obtain the exact solution. 3xsquare + 6x + 2 = 0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03x^2 + 6x + 2 = 0 You should use quadratic formula.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The problem I'm having is trying to find the correct square root of the problem.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4\[x ^{2}+2x+2/3\]\[3x ^{2}+6x+2=0\] dividing thru by 3. \[x ^{2}+2x + 1 +2/3 1=0\] \[(x+1)^{2}=1/3\]\[x+1=\sqrt{1/3}\] x+1=+/.577 x=1.577 x=.423

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks radar. You proved that I was doing it right, my book has the wrong answer in it.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4That happens sometimes (but seldom) Have you copied the problem correctly?

radar
 4 years ago
Best ResponseYou've already chosen the best response.4I'll enter the problem in the Wolfram machine to verify solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep. the answer they have in the book is\[3\pm \sqrt{3}\div3\] I tried doing 10 different ways and I couldn't get that But I got the answer you got every way I tried it

radar
 4 years ago
Best ResponseYou've already chosen the best response.4The Wolfram graphed the root very close to the values we have obtained. I would discuss it with one of your instructors. Good luck with your studies. I think you are correct.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4Looking at your book solution, it is similar to what we have:\[x+1 = 1/\sqrt{3}\]\[1/\sqrt{3}\times \sqrt{3}/\sqrt{3}=\sqrt{3}/3\]so in effect we have: \[x+1=\sqrt{3}/3\] or\[x=1\pm \sqrt{3}/3\] or\[3\pm \sqrt{3}\over 3\] yes Aurianaa, you have the equivalent of the book answer.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4It just looks different lol, but is the same.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4They rationalized the denominator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. I thought I was doing it wrong, This problem really upset me lol.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4No need to talk to your instructor. Or maybe discuss the fact the appearance of the book answer was confusing. I believe you are doing well.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4The problem was requiring you to use "completing the square" method. and as pratu043 suggested it could of been solved using the quadratic.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yea true I did that but I just couldn't figure out how to get the book answer when I knew I was doing it right.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4I hope all of this has not confused you because it appears that you have comprehension of the procedures.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4Do you see how the two answers are equivalent?

radar
 4 years ago
Best ResponseYou've already chosen the best response.4Do you see that multiplying that fraction by\[\sqrt{3}\over \sqrt{3}\] did not change its value only its appearance.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4Some people don't like radicals in the denominator, they rationalize them as your book did.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes i see how they can go both ways.

radar
 4 years ago
Best ResponseYou've already chosen the best response.4Good. time for my breakfast. good luck with your studies.
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