Answer :
To find the missing [tex]\(x\)[/tex]-value and its corresponding Pythagorean triple for the given problem, we proceed as follows:
Given side lengths are [tex]\(x^2 - 1\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x^2 + 1\)[/tex], and it is known that these form a right triangle.
1. Identify the missing [tex]\(x\)[/tex]-value and corresponding triple:
Let's check for an [tex]\(x\)[/tex]-value that gives a different triple from the ones provided in the question.
With [tex]\(x = 4\)[/tex], we calculate:
- [tex]\(t1 = x^2 - 1 = 4^2 - 1 = 16 - 1 = 15\)[/tex]
- [tex]\(t2 = 2x = 2 \cdot 4 = 8\)[/tex]
- [tex]\(t3 = x^2 + 1 = 4^2 + 1 = 16 + 1 = 17\)[/tex]
2. Sort the triple in ascending order:
The side lengths [tex]\(15\)[/tex], [tex]\(8\)[/tex], and [tex]\(17\)[/tex] are already in ascending order, so the Pythagorean triple is [tex]\( (8, 15, 17) \)[/tex].
3. Compile the results into the table:
The missing [tex]\(x\)[/tex]-value is [tex]\(4\)[/tex] and its corresponding triple is [tex]\( (8, 15, 17) \)[/tex].
Therefore, the completed table is:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex]-value & Pythagorean Triple \\
\hline
3 & (8, 15, 17) \\
\hline
5 & (8,15,17) \\
\hline
4 & (15,8,17) \\
\hline
\end{tabular}
Given side lengths are [tex]\(x^2 - 1\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x^2 + 1\)[/tex], and it is known that these form a right triangle.
1. Identify the missing [tex]\(x\)[/tex]-value and corresponding triple:
Let's check for an [tex]\(x\)[/tex]-value that gives a different triple from the ones provided in the question.
With [tex]\(x = 4\)[/tex], we calculate:
- [tex]\(t1 = x^2 - 1 = 4^2 - 1 = 16 - 1 = 15\)[/tex]
- [tex]\(t2 = 2x = 2 \cdot 4 = 8\)[/tex]
- [tex]\(t3 = x^2 + 1 = 4^2 + 1 = 16 + 1 = 17\)[/tex]
2. Sort the triple in ascending order:
The side lengths [tex]\(15\)[/tex], [tex]\(8\)[/tex], and [tex]\(17\)[/tex] are already in ascending order, so the Pythagorean triple is [tex]\( (8, 15, 17) \)[/tex].
3. Compile the results into the table:
The missing [tex]\(x\)[/tex]-value is [tex]\(4\)[/tex] and its corresponding triple is [tex]\( (8, 15, 17) \)[/tex].
Therefore, the completed table is:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex]-value & Pythagorean Triple \\
\hline
3 & (8, 15, 17) \\
\hline
5 & (8,15,17) \\
\hline
4 & (15,8,17) \\
\hline
\end{tabular}