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anonymous
 4 years ago
How do I solve by completing the square for: 1/2 n^2 + n = 1/8?
anonymous
 4 years ago
How do I solve by completing the square for: 1/2 n^2 + n = 1/8?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got n = 1 +/ √15/4 but the answer is supposed to be: 0.13, 3.87

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh, never mind I think I got it now!

radar
 4 years ago
Best ResponseYou've already chosen the best response.1Step 1. multiply the equation by 2 so that the coefficient of n^2 is 1, getting: n^2 + 2n = 2/8. O.K.

radar
 4 years ago
Best ResponseYou've already chosen the best response.1I'll stick around if you want to do the problem, I see you have figured it out.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so I'm getting confused: n + 1 = +/ √3/2 and n = 1 +/ √3/2

radar
 4 years ago
Best ResponseYou've already chosen the best response.1I will work it out and you can compare answers. Step 2 take half of the n coefficient (2) and square it and add it to the left side and right side. \[n ^{2}+2n + 1 = 2/8 +1\]\[(n + 1)^{2}=6/8\] We now have a perfect square on the left. Taking the square root of both sides give: n+1 =(6/8)^1/2 \[n+1 =\sqrt{6/8}\]\[n=1\pm \sqrt{6}/\sqrt{8}\] note that \[\sqrt{8}=2\sqrt{2}\]and\[\sqrt{6}=\sqrt{3}\sqrt{2}\] The fraction now becomes\[\sqrt{3}\over 2\] so the answer is:\[n =1\pm \sqrt{3}/2\]

radar
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, the fraction becomes simpler as it is reduced to sqrt3/2

radar
 4 years ago
Best ResponseYou've already chosen the best response.1Looks like we got the same answer. Good luck with these.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but.. isn't it wrong?

radar
 4 years ago
Best ResponseYou've already chosen the best response.1why do you think it is wrong?

radar
 4 years ago
Best ResponseYou've already chosen the best response.1Do you the answer as decimal numbers?
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