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April 30th, 2014, 03:27 PM  #1 
Newbie Joined: Apr 2014 From: dubai Posts: 1 Thanks: 0  Really hard algebra question
200y=x(x+40) Find x when y = 80 
April 30th, 2014, 04:03 PM  #2 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  $\displaystyle \color{blue} {y=80\ \Rightarrow\ 200\cdot80 = x(x+40)\\\;\\16 000 = x^2+40x_{+400} \ \Rightarrow\ 16400 = x^2+40x+400\\\;\\164\cdot100 = (x+20)^2\ \Rightarrow\ 41\cdot4\cdot100 = (x+20)^2\\\;\\(x+20)^2 = 4\cdot100\cdot41 \ \Rightarrow\ \sqrt{(x+20)^2} = \sqrt{4\cdot100\cdot41}\\\;\\x+20 = 2\cdot10\sqrt{41} \ \Rightarrow x+20 = 20\sqrt{41}\ \Rightarrow\ x+20 = \pm20\sqrt{41}\\\;\\x_1 = 2020\sqrt{41}\\\;\\x_2 = 20+20\sqrt{41}}$ 

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