The difference between two positive numbers is 3. If the sum of their squares is 369, then sum of the numbers is :

Question

Accepted Answer

The correct option is (2):
This problem requires the application of algebraic identities, specifically those relating the sum, difference, and sum of squares of two variables. Let the two positive numbers be $x$ and $y$. We are given two pieces of crucial information: first, the difference between the numbers is 3, which can be written as $x - y = 3$. Second, the sum of their squares is 369, expressed as $x^2 + y^2 = 369$. Our goal is to find the sum of the numbers, $x + y$.

The core concept for solving this lies in using the identity that connects these three components: the relationship between the squares of the sum and difference, and the sum of the squares itself. The identity is: $(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$. We can rearrange this formula to directly solve for the square of the sum, $(x + y)^2$.

Rearranging the identity gives:
$$(x + y)^2 = 2(x^2 + y^2) - (x - y)^2$$

Now, substitute the known values from the problem into this rearranged formula:
$$(x + y)^2 = 2(369) - (3)^2$$
$$(x + y)^2 = 738 - 9$$
$$(x + y)^2 = 729$$

To find the sum of the numbers ($x + y$), we must take the square root of 729. Since the problem specifies that the numbers are positive, their sum must also be positive.
$$x + y = \sqrt{729}$$
$$x + y = 27$$

Thus, the sum of the two positive numbers is 27, confirming that Option 2 is the correct answer. This method avoids the potentially complex process of solving the quadratic equations for $x$ and $y$ individually and relies instead on the efficiency of algebraic identities. (For context, if $x+y=27$ and $x-y=3$, solving simultaneously yields $x=15$ and $y=12$. Checking the sum of squares: $15^2 + 12^2 = 225 + 144 = 369$, which confirms the original conditions).

The incorrect option is (1):
Option 1 suggests that the sum of the numbers is 25. If $x+y=25$ and the difference is $x-y=3$, the two numbers would be 14 and 11. Checking the sum of their squares: $14^2 + 11^2 = 196 + 121 = 317$. Since the required sum of squares is 369, 25 is incorrect.

The incorrect option is (3):
Option 3 posits that the sum of the numbers is 31. If $x+y=31$ and $x-y=3$, the two numbers would be 17 and 14. Checking the sum of their squares: $17^2 + 14^2 = 289 + 196 = 485$. This value (485) is much greater than the given sum of squares (369), proving 31 is incorrect.

The incorrect option is (4):
Option 4 suggests the sum of the numbers is 33. If the sum is 33 and the difference is 3, the resulting numbers are 18 and 15. Checking the sum of their squares: $18^2 + 15^2 = 324 + 225 = 549$. Because the required sum of squares is 369, 33 is incorrect.