Question

Find two consecutive odd integers whose squares add up to 290.

Answer 1

Answer:

11, 13

Step-by-step explanation:

Let one of the odd positive integer be x , then the other odd positive integer is x+2.

Their sum of squares:

$x^{2} +(x+2)^{2}$

$x^{2} +(x+2)(x+2)$

$x^{2} +x^{2} +4x+4$

$2x^{2} +4x+4$

Given that their sum of squares = 290

$2x^{2} +4x+4 =290$

$2x^{2} +4x = 286$

$2x^{2} +4x -286 = 0$

$x^{2} +2x -143 =0$

$x^{2}+13x-11x-143 =0$

$x(x+13)-11(x+13)=0$

$(x+13)(x-11)=0$

Therefore x = 11, x=-13

We take positive value of x:

So, x =11 and (x+2)= 11+2=13

Therefore , the odd positive integers are 11 and 13