Question

Find two consecutive odd positive integers sum of whose squares is 290 10th clsss math

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

suppose that numbers be , x and (x+2)
a.t.q
x2 +(x+2)2 =290
solve it
x2 +2x -143 =0
by quadratic formula you can get
x = (-2+-24)/2
x = 11 and 13