Question

The product of two consecutive natural numbers is 3782.find the numbers. Write a equation to solve

Answer 1

Let the two consecutive natural numbers be \( n \) and \( n+1 \). The product of these two numbers can be expressed as:

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

Now, we can rearrange this equation:

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

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

Next, we can use the quadratic formula to find \( n \):

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

Plugging in the values we have:

\[
n = \frac{{-1 \pm \sqrt{{1^2 - 4 \times 1 \times (-3782)}}}}{{2 \times 1}}
\]

Calculating the discriminant:

\[
b^2 - 4ac = 1 + 4 \times 3782 = 1 + 15128 = 15129
\]

Now we can calculate \( \sqrt{15129} \):

\[
\sqrt{15129} = 123
\]

Now substituting back into the quadratic formula:

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

Calculating the two possible values for \( n \):

1. \( n = \frac{{122}}{{2}} = 61 \)
2. \( n = \frac{{-124}}{{2}} = -62 \) (not a natural number)

Thus, the only natural number solution is \( n = 61 \).

Therefore, the two consecutive natural numbers are:

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

To verify:

\[
61 \times 62 = 3782
\]

So, the solution is correct. The two consecutive natural numbers are \( 61 \) and \( 62 \).