## 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

1. anonymous

3x^2 + 6x + 2 = 0 You should use quadratic formula.

2. anonymous

The problem I'm having is trying to find the correct square root of the problem.

$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

4. anonymous

Thanks radar. You proved that I was doing it right, my book has the wrong answer in it.

That happens sometimes (but seldom) Have you copied the problem correctly?

I'll enter the problem in the Wolfram machine to verify solution.

7. 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

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.

9. anonymous

Thanks

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.

It just looks different lol, but is the same.

They rationalized the denominator.

13. anonymous

Ok. I thought I was doing it wrong, This problem really upset me lol.

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.

The problem was requiring you to use "completing the square" method. and as pratu043 suggested it could of been solved using the quadratic.

16. 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.

I hope all of this has not confused you because it appears that you have comprehension of the procedures.

Do you see how the two answers are equivalent?

Do you see that multiplying that fraction by$\sqrt{3}\over \sqrt{3}$ did not change its value only its appearance.

Some people don't like radicals in the denominator, they rationalize them as your book did.

21. anonymous

Yes i see how they can go both ways.