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What are two numbers for which the sum of there square is 628 and the differance of there squares is 340?
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x^2+y^2=628 x^2-y^2=360
solce to get the answer
a^2 + b^2 = 628 a^-b^2 = 340
let the numbers be a,b given \[a ^{2}+b ^{2}=628\]\[a ^{2}-b ^{2}=340\] then 2a^2=968 a^2=484 a=22 b=12
damnit Srir beat me :(
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who?me?
yeah you answered faster :P How did you know to add 340 to 628?
na.....its not the matter of speed its the matter of accuracy....see i'm not smart than anyone...so u too won!
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