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?

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
${\left(f + 1\right)}^{2} = {s}^{2}$
expanding
${f}^{2} + 2 \cdot f + 1 = {s}^{2}$ . . . . . (3)

Now (3) - (1) gives
$2 \cdot f - 22 = 0$
or $2 \cdot f = 22$
thus, $f = \frac{22}{2} = 11$
and $s = f + 1 = 11 + 1 = 12$
So the numbers are 11 & 12