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anonymous
 one year ago
If a square has an area of 4x^22xy +49y^2 sq cm.
Find the length of its side and solve its perimeter.
Please help, am I going to use 'completing the square'?
Thanks!
anonymous
 one year ago
If a square has an area of 4x^22xy +49y^2 sq cm. Find the length of its side and solve its perimeter. Please help, am I going to use 'completing the square'? Thanks!

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phi
 one year ago
Best ResponseYou've already chosen the best response.2the area of a square is s*s (where s is the length of its side) that is, s^2 so I would expect the expression 4x^22xy +49y^2 to be a perfect square are you sure it is really 2xy for the middle term ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2in that case, I would just say the side has length \[ s = \sqrt{4x^22xy +49y^2} \] and the perimeter is 4s, so \[ 4\sqrt{4x^22xy +49y^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh, okay2 thanks! :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if x =1, y =2 we have the expression = 14^2 , but not sure how to go further.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^ yeah :( thank you though!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, I find out y =2x works for all the case since if we replace y =2x, the first term and the second term cancel out 4x^2 2(x)(2x) =4x^2 4x^2 =0 while the last term is always a perfect square since 49 y^2 = (7y)^2 Hence for all couple (x,2x) we have a square satisfy the condition.
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