Question

The sum of two natural numbers 45. The product of these numbers is four times the square of the lesser number. Find the numbers.

Answer 1

Let's assume that the two natural numbers are x and y, where x is the smaller number.

From the problem, we know that:

x + y = 45 ... (1) (The sum of the two numbers is 45)

xy = 4x^2 ... (2) (The product of the two numbers is four times the square of the lesser number)

We can rearrange equation (1) to get:

y = 45 - x

Substituting this into equation (2), we get:

x(45 - x) = 4x^2

Expanding the left side and simplifying, we get:

45x - x^2 = 4x^2

Rearranging and simplifying further, we get a quadratic equation:

5x^2 - 45x = 0

We can factor out x to get:

x(5x - 45) = 0

So, either x = 0 (which is not a natural number) or 5x - 45 = 0.

Solving for x, we get:

5x - 45 = 0

5x = 45

x = 9

So the smaller number is 9, and using equation (1), we can find that the larger number is:

y = 45 - x = 45 - 9 = 36

Therefore, the two natural numbers are 9 and 36.

Answer 2

Explanation:

Let's call the two natural numbers x and y. We know that:

x + y = 45 (Equation 1)

xy = 4x^2 (Equation 2)

We can solve Equation 1 for y:

y = 45 - x

Then, we can substitute this into Equation 2:

x(45 - x) = 4x^2

Expanding and simplifying:

45x - x^2 = 4x^2

5x^2 - 45x = 0

We can factor out x:

x(5x - 45) = 0

This gives us two possible solutions: x = 0 or x = 9. Since we are dealing with natural numbers, x cannot be 0. Therefore, x = 9.

We can substitute this back into Equation 1 to find y:

9 + y = 45

y = 36

So the two natural numbers are 9 and 36.