Question

The sum of two natural numbers is 21 and the sum of their squares is 233 find the numbers

Answer 1

let first number be x then other number will be
(21-x). as per given
x^2 + (21-x)^2= 233
x^2 + 441 + x^2 - 42x = 233
2x^2 - 42x + 441 -233 =0
2x^2 -42x + 208 = 0
dividing by on both sides
x^2 -21x +104 =0
x^2 - 8x - 13x +104 =0
x(x-8) -13(x-8) =0
(x-8) (x-13)=0
therefore x = 8 or 13
then the numbers are x=8
and 21-x = 21-8= 13.

hope it helps.....,

Answer 2

Answer:

Let no x^2

OTHE NO (X-12)^2

AC+Q

X^2 +X^2 - 42 +441 =233

2X^2 - 42X +208

X(X-8)-13(X-8)=0

(X-8)(X-13)

X=8

X=13