The product of two consecutive natural numbers is 3782. Find the number

Question

Could you solve for these word problem

Answer 1

Let the two consecutive natural numbers be \( n \) and \( n + 1 \). We know that their product is given by:

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

This can be rewritten as:

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

Now, we can solve this quadratic equation using the quadratic formula:

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

Here, \( a = 1\), \( b = 1\), and \( c = -3782\).

Calculating the discriminant:

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

Now, take the square root of the discriminant:

\[ \sqrt{15129} = 123 \]

Now we can substitute back into the quadratic formula:

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

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

1. \( n = \frac{{-1 + 123}}{2} = \frac{122}{2} = 61 \)
2. \( n = \frac{{-1 - 123}}{2} = \frac{-124}{2} = -62 \) (not a natural number)

Thus, the only solution is \( n = 61 \). Therefore, the two consecutive natural numbers are:

\[ 61 \text{ and } 62 \]

To confirm, we can check the product:

\[ 61 \times 62 = 3782 \]

So, the two consecutive natural numbers are \( 61 \) and \( 62 \).

Answer 2

What are consecutive natural numbers?

Answer 3

Consecutive natural numbers are numbers that follow each other in order without any gaps. In other words, they are natural numbers that are one more than the previous number. For example, the first few consecutive natural numbers are:

• 1 and 2
• 2 and 3
• 3 and 4
• 4 and 5

In general, if \( n \) is a natural number, then the consecutive natural numbers are \( n \) and \( n + 1 \).

Natural numbers themselves are the set of positive integers starting from 1: {1, 2, 3, 4, 5, ...}. So, any pair of numbers in this set that are next to each other is considered a pair of consecutive natural numbers.

Answer 4

Could you simplify the first answer?

Answer 5

Sure! Here’s a simplified version of the explanation for finding the two consecutive natural numbers whose product is 3782:

1. Define the numbers: Let the two consecutive natural numbers be \( n \) and \( n + 1 \).

2. Set up the equation: Their product can be expressed as: \[ n(n + 1) = 3782 \]

3. Rearrange the equation: This can be rewritten as: \[ n^2 + n - 3782 = 0 \]

4. Use the quadratic formula: The quadratic formula \( n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) applies here, with \( a = 1 \), \( b = 1 \), and \( c = -3782 \).

   • Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-3782) = 1 + 15128 = 15129 \]
   • Take the square root: \[ \sqrt{15129} = 123 \]

5. Find the values of \( n \): \[ n = \frac{{-1 \pm 123}}{2} \]

   • The solutions are:
     • \( n = \frac{122}{2} = 61 \) (valid)
     • \( n = \frac{-124}{2} = -62 \) (not a natural number)

6. Conclude the results: The two consecutive natural numbers are: \[ 61 \text{ and } 62 \]

Thus, the final answer is that the two consecutive natural numbers whose product is 3782 are 61 and 62.

Answer 6

Rewrite the quadratic formula again to a more simple form?
Please

Answer 7

Certainly! The quadratic formula is used to solve equations of the form:

\[ ax^2 + bx + c = 0 \]

The formula to find the solutions (roots) for \( x \) is:

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

To simplify the understanding, you can express it in a more intuitive way:

1. Identify the coefficients: \( a \), \( b \), and \( c \) are numbers from your quadratic equation.

2. Discriminant: \( D = b^2 - 4ac \) determines the nature of the roots:

   • If \( D > 0 \): two real and distinct roots.
   • If \( D = 0 \): one real root (the roots are the same).
   • If \( D < 0 \): no real roots (the roots are complex).

3. Final solution: The solutions for \( x \) can be written as: \[ x = \frac{{-b \pm \sqrt{D}}}{2a} \]

So, in its simpler form, the quadratic formula gives you two possible values for \( x \) based on whether you add or subtract the square root of the discriminant from \(-b\) and then divide by \(2a\).

Example: If you have an equation like \( x^2 + 2x - 8 = 0 \) (where \( a = 1 \), \( b = 2 \), \( c = -8 \)), you would calculate:

1. Discriminant: \( D = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \)
2. Solutions: \[ x = \frac{{-2 \pm \sqrt{36}}}{2 \cdot 1} = \frac{{-2 \pm 6}}{2} \] Which gives you:
   • \( x = \frac{4}{2} = 2 \)
   • \( x = \frac{-8}{2} = -4 \)

I hope this makes it clearer! Let me know if you need more information.