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July 17th, 2014, 06:52 AM  #1 
Newbie Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0  Year 11 Completing the Square Question
Question: x^2 + 6x = 11 How would I solve for x? The answer at the back of the book is: x = 3 +/ √11 Help me please! Its for my maths homework. I believe the solution is found by using completing the square because the example above the question uses this method: The example was: x^2 + 4x + 1 = 0 the answer is: x = 2 +/ √3 Last edited by wrightarya; July 17th, 2014 at 07:03 AM. Reason: additional info that could be helpful to solve problem 
July 17th, 2014, 08:05 AM  #2 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176  $x^2+6x+11=0 \\\;\\ x_{1,2}=\dfrac{b\pm\sqrt{\Delta}}{2a} \\\;\\ \Delta = b^24ac. \\\;\\ To \ further, \ with \ care\ caution, \ it\ is \ dry\ sand \\\;\\ (Analog,\ for \\\;\\ \ x^2\ +\ 6x\ \ 11\ =\ 0\ \ ! \ )$ 
July 17th, 2014, 09:49 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,912 Thanks: 883  
July 17th, 2014, 06:31 PM  #4 
Newbie Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0 
I triple checked the question and answer in the textbook. I haven't wrote it down incorrectly. I substituted the answer in for x and I see how it the question should be x^ + 6x = 2 maybe the Haese year 11 textbook made a misprint! 
July 17th, 2014, 11:43 PM  #5 
Newbie Joined: Jul 2014 From: Australia Posts: 24 Thanks: 0 
The solution for the example question is: x^2 + 4x + 1 = 0 x^2 + 4x + 2^2 + 1  2^2 = 0 (x+2)^2 3 = 0 (x+2)^2 = 3 x+2 = √3 x= 2 +/ √3 That's how they solved it in the book. But this method doesn't seem to work for the question I posted which was this question: x^2 + 6x = 11 Last edited by wrightarya; July 18th, 2014 at 12:08 AM. 
July 17th, 2014, 11:57 PM  #6 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176  
July 18th, 2014, 12:13 AM  #7 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176  

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algebra, completing, quadratic, question, square, year, year 11 
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