The square of one number is 23 less than the square of a second number. If the second number is 1 more than the first, what are the two numbers?

1 Answer
Mar 11, 2016

The numbers are 11 & 12

Explanation:

Let the first number be f and the second |number be s
Now the square of first No. is 23 less than the square of second No.
ie. f^2 + 23 = s^2 . . . . . (1)
The second No. is 1 more than the first
ie f + 1 = s . . . . . . . . . . .(2)
squaring (2), we get
(f + 1)^2 = s^2
expanding
f^2 + 2 * f + 1 = s^2 . . . . . (3)

Now (3) - (1) gives
2 * f - 22 = 0
or 2 *f = 22
thus, f = 22 / 2 = 11
and s = f + 1 = 11 + 1 = 12
So the numbers are 11 & 12