Question

Three consecutive natural numbers are such that the square of the middle number excceds the difference of the squares of the other two by 60.

Assume the middle number to be x and form a quadratic equation satisfying's the above statement. Hence; find the three numbers.

Answer 1

Let the numbers be x – 1, x and x + 1.
From the given information,
$x^2=(x+1{)}^2–(x–1{)}^2+60$
$x^2=x^2+1+2x–x^2–1+2x+60$
$x^2=4x+60$
$x^2–4x–60=0$
(x – 10) (x + 6) = 0
x = 10, -6
Since, x is a natural number, so x = 10.
Thus, the three numbers are 9, 10 and 11.