Question

Two successive odd integers are such that the sum of their square is 290. Find the integers.

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

odd integer is (2x+1) and (2x+3) .

=>(2x+1)² + (2x+3)² = 290

=>4x²+ 4x + 1 +4x² +12x + 9= 290

=> 8x² + 16x + 10 -290 = 0

=> 8x² + 16x - 280=0

=> 8 ( x² + 2x - 35 )= 0

=> 8 ×( x² + 7x - 5x -35 )=0

=> 8 [ x ( x +7 ) - 5 ( x + 7)]=0

=> 8 . (x-5) ( x +7 ) =0

=> therefore , x = 5 and -7
now , integer = 2x +1 = (11 and 13)

and =( -13 and -11)