Question

Find the sum of two consecutive positive integers, such that the sum of their squares is 365.
[3 Marks]

Answer 1

Let the two consecutive positive integers be $x\&x+1.$
[0.5 Marks]

Given, the sum of squares of the positive consecutive integers is 365.
$\therefore x^2+(x+1{)}^2=365$ [0.5 Marks]

Lets solve
$x^2+x^2+2x+1=3652x^2+2x-364=0x^2+x-182=0$
(Dividing by 2)
​​​​​​​[0.5 Marks]

Factorise,
$x^2+14x-13x-182=0x(x+14)-13(x+14)=0(x+14)(x-13)=0\therefore x=13\&-14$
​​​​​​​[0.5 Marks]

Since, the given numbers are positive integers,
hence $x=13$.

So, the required consecutive numbers are $13\&14.$ ​​​​​​​[0.5 Marks]
and thus the sum of the given consecutive positive integers = 13 + 14 = 27 ​​​​​​​[0.5 Marks]