Question:
We know that a triangle with side lengths [tex]\(x^2 - 1\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x^2 + 1\)[/tex] is a right triangle. Using those side lengths, find the missing triples and [tex]\(x\)[/tex]-values.

Write the triples in parentheses, without spaces between the numbers, and with a comma between numbers. Write the triples in order from least to greatest.

Type the correct answer in each box.

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$-value & Pythagorean Triple \\
\hline
3 & $(8,15,17)$ \\
\hline
5 & $(20,21,29)$ \\
\hline
$\square$ & $\square$ \\
\hline
$\square$ & $\square$ \\
\hline
$\square$ & $\square$ \\
\hline
\end{tabular}
\][/tex]



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}