Question

The product of two consecutive natural numbers is 3782. Find the numbers?
Solve for these?

Answer 1

Let the two consecutive natural numbers be \( n \) and \( n + 1 \). The equation representing their product can be written as:

\[
n(n + 1) = 3782
\]

Expanding this, we have:

\[
n^2 + n - 3782 = 0
\]

This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 1 \), and \( c = -3782 \).

To find the values of \( n \), we can use the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Plugging in the values:

\[
b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-3782) = 1 + 15128 = 15129
\]

Now, calculating the square root:

\[
\sqrt{15129} = 123
\]

Now, substituting back into the quadratic formula:

\[
n = \frac{-1 \pm 123}{2}
\]

Calculating the two possible values:

1.

\[
n = \frac{122}{2} = 61
\]

2.

\[
n = \frac{-124}{2} = -62 \quad (\text{not a natural number})
\]

The only valid solution is \( n = 61 \).

Thus, the two consecutive natural numbers are:

\[
61 \quad \text{and} \quad 62
\]

To check:

\[
61 \times 62 = 3782
\]

Therefore, the solution to the problem is:

**The two consecutive natural numbers are 61 and 62.**