Question

If the sum of the square of two consecutive odd positive integers is 290 find the numbers

Answer 1

Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares = x² +(x+2)²
= x² + x² + 4x +4
= 2x² + 4x + 4
Given that their sum of squares = 290
⇒ 2x² +4x + 4 = 290
⇒ 2x² +4x = 290-4 = 286
⇒ 2x² + 4x -286 = 0
⇒ 2(x² + 2x - 143) = 0
⇒ x² + 2x - 143 = 0
⇒ x² + 13x - 11x -143 = 0
⇒ x(x+13) - 11(x+13) = 0
⇒ (x-11) = 0 , (x+13) = 0
Therfore , x = 11 or -13
We always take positive value of x
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .

Answer 2

let the two odd numbers be A and (A+1),

then according to the question, we have

A²+(A+2)²=290,

A²+A²+4+4A=290,

then

2A²+4A=286,

A²+2A-143=0,

then

A²+(13-11)A-143=0,

A²+13A-11A-143=0,

A(A+13)-11(A+13)=0,

(A+13)(A-11)=0,

if

A-11=0,

then

A=11,

therefore

first odd number=11,

second odd number=11+2=13