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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

13, 14

-13,-14

-13,14

13,-14

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let the first integer be x.
The next consecutive positive integer will be x + 1.
According to the given question,
x² + ( x + 1)² = 365
Or, x² +( x + 1)² = 365
Or, x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]
Or, 2x² + 2x + 1 = 365
Or, 2x² + 2x + 1- 365 = 0
Or, 2x² + 2x - 364 = 0
Or, 2(x² + x - 182) = 0
Or, x² + x - 182 = 0
Or, x² + 14x - 13x - 182 = 0
Or, x (x + 14) - 13 (x + 14) = 0
Or, (x - 13) (x + 14) = 0
Or, x - 13 = 0 and x + 14 = 0
x = 13 and x = - 14
The value of x cannot be negative (because it is given that the integers are positive).
∴ x = 13 and x + 1 = 14
Hence, Option 1 is the correct option.

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th