Sure! Here is the detailed step-by-step solution:
Given the side lengths of a right triangle as [tex]\(x^2 - 1\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x^2 + 1\)[/tex]:
1. When [tex]\(x = 3\)[/tex]:
- [tex]\(x^2 - 1 = 3^2 - 1 = 9 - 1 = 8\)[/tex]
- [tex]\(2x = 2 \times 3 = 6\)[/tex]
- [tex]\(x^2 + 1 = 3^2 + 1 = 9 + 1 = 10\)[/tex]
So, the triple is [tex]\((6, 8, 10)\)[/tex].
2. When [tex]\(x = 5\)[/tex]:
- [tex]\(x^2 - 1 = 5^2 - 1 = 25 - 1 = 24\)[/tex]
- [tex]\(2x = 2 \times 5 = 10\)[/tex]
- [tex]\(x^2 + 1 = 5^2 + 1 = 25 + 1 = 26\)[/tex]
So, the triple is [tex]\((10, 24, 26)\)[/tex].
3. When [tex]\(x = 6\)[/tex]:
- [tex]\(x^2 - 1 = 6^2 - 1 = 36 - 1 = 35\)[/tex]
- [tex]\(2x = 2 \times 6 = 12\)[/tex]
- [tex]\(x^2 + 1 = 6^2 + 1 = 36 + 1 = 37\)[/tex]
So, the triple is [tex]\((12, 35, 37)\)[/tex].
Hence the table would be filled as follows:
\begin{tabular}{|c|c|}
\hline[tex]$x$[/tex]-value & Pythagorean Triple \\
\hline 3 & (6,8,10) \\
\hline 5 & (10,24,26) \\
\hline 6 & (12,35,37) \\
\hline
\end{tabular}
