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Find two consecutive positive integers, sum of whose squares is 365.
Dear StudentLet us say, the two consecutive positive integers bexandx+ 1.Therefore, as per the given questions,x^2+ (x+ 1)^2= 365⇒x^2+x^2+ 1 + 2x= 365⇒2x^2+ 2x - 364 = 0⇒x^2+x- 182 = 0⇒x^2+ 14x- 13x- 182 = 0⇒x(x+ 14) -13(x+ 14) = 0⇒(x+ 14)(x- 13) = 0Thus, either,x+ 14 = 0 orx- 13 = 0,⇒x= - 14 orx= 13since, the integers are positive, soxcan be 13, only.x+ 1 = 13 + 1 = 14 two consecutive positive integers will be 13 and 14.Thanks
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