Question

Find two consecutive positive integers ,sum of whose square is 365 find the number

Answer 1

Let The Numbers Be x And x+1

Squares x2 And (x+1)2

Sum = 365

Hence

x2+(x+1)2 = 365

x2+x2+1+2x = 365

2x2+2x+1 = 365

2x2+2x-364 = 0

x2+x-182 = 0

Use Quadratic Formula

-b+Square Root (b2-4ac) / 2a    And -b-Square Root (b2-4ac) / 2a

-1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2

-1+27/2 And -1-27/2

13 And -14

As Positive Integers So 13

Other Integer = 13+1 = 14

Answer 2

Let the two consecutive positive integers be x and x + 1.

GIven that sum of whose squares = 365.

x^2 + (x + 1)^2 = 365

x^2 + x^2 + 1 + 2x = 365

2x^2 + 2x + 1 = 365

2x^2 + 2x = 364

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

x = -14 (or) x = 13.


Since x cannot be -ve. So,x = 13.

Then, x + 1 = 14.


Therefore the numbers are 13 and 14.


Hope this helps!