Question 75016
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This is the second problem where you have ended up with an extra multiplier of 2. Also
you forgot the common term in front of the radical and you got the wrong value under the 
radical.  Let's see if we can clear up things by working the example step-by-step. 
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We'll complete the square and maybe you can see where that 2 is coming from along with
correcting the other mistakes:
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{{{x^2-6x-3=0}}}
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The first 2 terms are {{{x^2 - 6x}}} Let's make this a perfect square.  Divide the -6 by 2 to get
-3.  Then square the -3 to get + 9.  So we add and subtract 9 from the equation as follows:
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{{{x^2-6x+9 -9 -3=0}}}
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Combine the -9 and the -3 to get -12. This makes the equation:
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{{{x^2-6x+ 9 -12=0}}}
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Add 12 to both sides of the equation and you have:
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{{{x^2-6x+9=12}}} 
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Since the left side of the equation is a perfect square, we can write it as:
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{{{(x - 3)^2 = 12}}}
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Take the square root of both sides:
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{{{x-3 = sqrt(12) = sqrt(4*3) = sqrt(4)*sqrt(3)= 2sqrt(3)}}}
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When you take the sqrt(3) it involves both plus and minus signs.  So the problem will have 2
answers ... one with a + sign and one with a - sign.
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The equation is now:
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{{{x - 3 = 2sqrt(3)}}}
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Add 3 to both sides to solve for x and the two answers become:
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{{{x = 3 + 2sqrt(3)}}} and {{{x = 3 - 2sqrt(3)}}}
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Hope this helps you identify and clarify the problems that popped up in your answer.