Question

find the two consecutive odd positive integers, sum of whose
squares are 290.​

Answer 1

11&13

121+169

u can choose many different methods to find it

let x, x+2

x^2+x^2+2x+4=290

2x^2+2x-286=0

2x^2+26x-22x-286=0

..

find itx=11

Answer 2

Answer:

Let numbers x and x+2

${x}^{2} + {(x + 2)}^{2} = 290 \\ {x}^{2} + {x}^{2} + 4x + 4= 290 \\ 2 {x}^{2} + 4x - 286 = 0 \\ 2( {x}^{2} + 2x - 143) = 0 \\ 2( {x}^{2} + 13x - 11x - 11 \times 13) = 0 \\ 2(x + 13)(x - 11) = 0 \\ since \: \: x \: \: is \: \: positive \\ x = 11$

And other will be 11 + 2 = 13