Question

Encuentra dos numeros positivos tal que su cuadrado menos su quintuplo sea 36

Answer 1

The two positive numbers are:
`\[ (x, y) = \left((5 + √(221))/(2), (7 + √(221))/(2)\right) \]`

Let's denote the two positive numbers as
`\(x\)` and
`\(y\)`. The problem states that their square minus their fivefold is equal to 36. We can express this algebraically:

`\[ x^2 - 5y = 36 \]`

Now, we need another relationship between
`\(x\)` and
`\(y\)`. Since the problem doesn't provide one explicitly, let's assume another relationship that might make the problem solvable. For example, we can assume that the two numbers are consecutive. Therefore:

`\[ y = x + 1 \]`

Now, we can substitute
`\(y = x + 1\)` into the first equation:

`\[ x^2 - 5(x + 1) = 36 \]`

Now, solve this equation to find the values of
`\(x\)` and
`\(y\)`:

`\[ x^2 - 5x - 41 = 0 \]`

Using the quadratic formula:

`\[ x = (-b \pm √(b^2 - 4ac))/(2a) \]`

For the equation
`\(x^2 - 5x - 41 = 0\)`,
`\(a = 1\)`,
`\(b = -5\)`, and
`\(c = -41\)`. Plugging these values into the quadratic formula:

`\[ x = (5 \pm √(5^2 - 4(1)(-41)))/(2(1)) \]`

`\[ x = (5 \pm √(221))/(2) \]`

The square root of 221 is an irrational number, so there are two solutions for
`\(x\)`, one using the positive square root and one using the negative square root.

The two positive numbers are:

`\[ (x, y) = \left((5 + √(221))/(2), (7 + √(221))/(2)\right) \]`

And the other solution using the negative square root:

`\[ (x, y) = \left((5 - √(221))/(2), (7 - √(221))/(2)\right) \]`

Both solutions satisfy the conditions given in the problem.