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April 30th, 2014, 03:27 PM   #1
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Really hard algebra question

200y=x(x+40)

Find x when y = 80
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April 30th, 2014, 04:03 PM   #2
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Quote:
Originally Posted by beerato View Post
200y=x(x+40)

Find x when y = 80

$\displaystyle \color{blue} {y=80\ \Rightarrow\ 200\cdot80 = x(x+40)\\\;\\16 000 = x^2+40x|_{+400} \ \Rightarrow\ 16400 = x^2+40x+400\\\;\\164\cdot100 = (x+20)^2\ \Rightarrow\ 41\cdot4\cdot100 = (x+20)^2\\\;\\(x+20)^2 = 4\cdot100\cdot41 \ \Rightarrow\ \sqrt{(x+20)^2} = \sqrt{4\cdot100\cdot41}\\\;\\|x+20| = 2\cdot10\sqrt{41} \ \Rightarrow |x+20| = 20\sqrt{41}\ \Rightarrow\ x+20 = \pm20\sqrt{41}\\\;\\x_1 = -20-20\sqrt{41}\\\;\\x_2 = -20+20\sqrt{41}}$
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