anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
3x^2 + 6x + 2 = 0 You should use quadratic formula.
anonymous
  • anonymous
The problem I'm having is trying to find the correct square root of the problem.
radar
  • radar
\[x ^{2}+2x+2/3\]\[3x ^{2}+6x+2=0\] 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

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anonymous
  • anonymous
Thanks radar. You proved that I was doing it right, my book has the wrong answer in it.
radar
  • radar
That happens sometimes (but seldom) Have you copied the problem correctly?
radar
  • radar
I'll enter the problem in the Wolfram machine to verify solution.
anonymous
  • anonymous
Yep. 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
  • radar
The 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.
anonymous
  • anonymous
Thanks
radar
  • radar
Looking 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
  • radar
It just looks different lol, but is the same.
radar
  • radar
They rationalized the denominator.
anonymous
  • anonymous
Ok. I thought I was doing it wrong, This problem really upset me lol.
radar
  • radar
No 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
  • radar
The problem was requiring you to use "completing the square" method. and as pratu043 suggested it could of been solved using the quadratic.
anonymous
  • anonymous
Yea 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
  • radar
I hope all of this has not confused you because it appears that you have comprehension of the procedures.
radar
  • radar
Do you see how the two answers are equivalent?
radar
  • radar
Do you see that multiplying that fraction by\[\sqrt{3}\over \sqrt{3}\] did not change its value only its appearance.
radar
  • radar
Some people don't like radicals in the denominator, they rationalize them as your book did.
anonymous
  • anonymous
Yes i see how they can go both ways.
radar
  • radar
Good. time for my breakfast. good luck with your studies.

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