Ex 4.2, 4
Find two consecutive positive integers, sum of whose squares is 365.
There is difference of 1 in consecutive positive integers
Let First integer = x
∴ Second integer = x + 1
Given that
Sum of squares = 365
(First number)2 + (Second number)2 = 365
x2 + (x + 1)2 = 365
x2 + x2 + 12 + 2 × x ×1=365
2x2 + 1 + 2x = 365
x2 + x – 182 = 0/2
2x2 + 2x +1 – 365 = 0
2x2 + 2x – 364 = 0
2 (x2 + x – 182) = 0
x2 + x – 182 = 0
We factorize by
splitting the middle term method
x2 + 14x – 13x – 182 = 0
x (x + 14) – 13 (x + 14) = 0
(x – 13) (x + 14) = 0
So, the root of the equation are x = 13 & x = – 14
x – 13 = 0
x = 13
x + 14 = 0
x = –14
Since we have to find consecutive positive numbers
x = −14 is not possible
Thus, x = 13
Therefore,
First number = x = 13
Second number = x + 1 = 13 + 1 = 14

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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