Question

solve the quadratic equations:

the sum of the squares of two consecutive natural odd number is 802. find them​

Answer 1

Let one odd number be X

the other odd number will be X+2

A/Q

X^2+(X+2)^=802

X^2+X^2+4X+4=802

2X^2+4X-798=0

X^2+2X-399=0

(X+21)(X-19)=0

X=-21(REJECTED)

therefore

X=19

SO OTHER ODD NUMBER IS 21