Alg3.2.19

Find two consecutive odd natural numbers the sum of whose squares is 202

Let the first odd number be $x\,$

Then,the second odd number is $(x+2)\,$

Sum of the squares of the two numbers is $x^2+(x+2)^2\,$

Given

$x^2+(x+2)^2=202\,$

$x^2+x^2+4x+4=202,2x^2+4x-198=0,x^2+2x-99=0\,$

Solving the quadratic equation,we get

$x^2+11x-9x-99=0,x(x+11)-9(x+11)=0,(x+11)(x-9)=0\,$

Hence $x=-11,9\,$

Since $x\,$ is a natural number,the value is $9\,$

Therefore,the two consecutive odd natural numbers are $9,11\,$