find two consecutive positive integers, sum of whose squares is 365
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(13×13 +14×14=365
13 and14 are consequetive
13 and14 are consequetive
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let first integer be x
they are consecutive then other will be x+1
x^2 + (x+1)^2 = 365
x^2 + x^2 + 1 + 2x =365
2x^2 + 2x = 365-1
2(x^2 + x ) = 364
x^2 + x = 182
x^2 + x - 182 = 0
x^2 + 14x - 13x - 182 = 0
x(x+14) -13(x+14) = 0
(x+14) (x-13) = 0
x+14=0 ; x-13=0
x= -14 ; x=13
x could not be negative
x=13
then integers are 13 , 13 + 1 = 14
hope it help you please like it
they are consecutive then other will be x+1
x^2 + (x+1)^2 = 365
x^2 + x^2 + 1 + 2x =365
2x^2 + 2x = 365-1
2(x^2 + x ) = 364
x^2 + x = 182
x^2 + x - 182 = 0
x^2 + 14x - 13x - 182 = 0
x(x+14) -13(x+14) = 0
(x+14) (x-13) = 0
x+14=0 ; x-13=0
x= -14 ; x=13
x could not be negative
x=13
then integers are 13 , 13 + 1 = 14
hope it help you please like it
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