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

Question

Harshit Singh · Answer

Dear Student

Let 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) = 0

Thus, either,x+ 14 = 0 orx- 13 = 0,

⇒x= - 14 orx= 13
since, the integers are positive, soxcan be 13, only.

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

Thanks

Pankaj Suyal · Answer

Let the first no. = X Second no. = X+1 (consecutive number mean continuously) According to the question (X) 2 +(X+1) 2 = 365 X 2 +X 2 +1+2X = 365 2X 2 +2X=365-1 2X 2 +2X=364 Dividing both sides by 2 X 2 +X=182 X 2 +X-182=0 Now cwe factorize this equation X 2 +14X-13X-182=0 X(X+14)-13(X+14)=0 (x+14)(x-13)=0 X+14=0 x=-14 x-13=0 x=13 Given in question positive integer so we select 13 which is value of x so can say that X=13 X+1=13+1=14 Thank you 2X 2+ 2X+1=365

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

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

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

2 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on 10 grade maths

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship