Sorry to correct you but I only know (m2-1),(2m)

Thank you for clarifying. If we are to find a Pythagorean triplet where one of its members is 15 using the formulas \((m^2 - 1)\) and \((2m)\), then we can follow these steps:

### Formula
The formulas given are:
- \(a = m^2 - 1\)
- \(b = 2m\)
- \(c = m^2 + 1\)

We need to find \(m\) such that either \(a\) or \(b\) equals 15.

### Case 1: \(a = 15\)
\[ m^2 - 1 = 15 \]

Solving for \(m\):
\[ m^2 = 16 \implies m = 4 \]

Then:
- \( a = m^2 - 1 = 4^2 - 1 = 16 - 1 = 15 \)
- \( b = 2m = 2 \times 4 = 8 \)
- \( c = m^2 + 1 = 4^2 + 1 = 16 + 1 = 17 \)

Thus, the triplet is \((15, 8, 17)\).

### Case 2: \(b = 15\)
\[ 2m = 15 \]

Solving for \(m\):
\[ m = 7.5 \]

Since \(m\) must be an integer, \( b \neq 15 \) is not possible.

### Conclusion
The valid Pythagorean triplet where one member is 15, using the given formulas, is \((15, 8, 17)\).

Would you like more details on the calculations or have any questions?

Here are some questions you might want to ask:
1. Can you explain the derivation of the formulas \(a = m^2 - 1\) and \(b = 2m\)?
2. What are some other common formulas to generate Pythagorean triplets?
3. How do you verify if a set of numbers forms a Pythagorean triplet?
4. Are there any other methods to find Pythagorean triplets?
5. How do Pythagorean triplets apply to real-world problems?

**Tip:** When using formulas to find specific values, it's crucial to ensure all conditions (like integer constraints) are met for the solutions to be valid.

Explore how to find Pythagorean triplets using the formulas m2-1 and 2m. Step-by-step solution provided for a problem involving a Pythagorean triplet where one member is 15. Suitable for high school students studying algebra and the Pythagorean theorem.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation