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

Answer:

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