From the question we are to find two natural numbers which differ by 3, such that the sum of their squares is 117.
Let the numbers b x and y
Given that:
x - y = 3 ......................... eqn I
x² + y² = 117 .................. eqn II
From eqn I: x - y = 3
Squaring both sides:
(x - y)² = 3²
(x - y)² = 9
x² + b² - 2xy = 9
substitute the value of x² + y² from eqn II in eqn I
117 - 2xy = 9
2xy = 108
divide both sides by 2
xy = 108/2
xy = 54 .......................... eqn III
divide both sides by a
y = 54/x
substitute the above value of y in eqn I:
x - 54/x = 3
x² - 54 = 3x
x² - 3x - 54 = 0
solving the quadratic:
x = -(-3) ± √(-3)² - 4 * 1 * (-54)
2 * 1
x = √(-3)² - 4 * 1 * (-54) = 15
x = -(-3) ± 15
2 * 1
x = 9, -6
but -6 is not a solution because it is given that they are natural numbers:
So, x = 9
inputting x = 9 in eqn I
x - y = 3
9 - y = 3
y = 9 - 3
y = 6
So,
x = 9
y = 6
Therefore, the two natural numbers are: 9 & 6.