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 \( n \):
- \( n = \frac{{122}}{{2}} = 61 \)
- \( n = \frac{{-124}}{{2}} = -62 \) (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 \( 62 \).
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