Question

Find the two natural numbers which differ by 5 and the sum of whose squares is 97.

Answer 1

Let the two numbers be x and x + 5.
From the given information,
$x^2+(x+5{)}^2=97$
$2\times x^2+10\times x+25–97=0$
$2\times x^2+10\times x–72=0$
$x^2+5\times x–36=0$
(x + 9) (x – 4) = 0
x = -9 or 4
Since, -9 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 9.