Find two consecutive positive integers, sum of whose squares is 365. Q.4

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The important question of class 10th of quadratic equations, how i solve the question of exercise 4.2 of question no.4 easily Find two consecutive positive integers, sum of whose squares is 365.

Let us say, the two consecutive positive integers be

xandx+ 1.Therefore, as per the given questions,

x^{2}+ (x+ 1)^{2}= 365⇒

x^{2 }+x^{2 }+ 1 + 2x= 365⇒ 2

x^{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, so

xcan be 13, only.∴

x+ 1 = 13 + 1 = 14Therefore, two consecutive positive integers will be 13 and 14.

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