To solve for the missing triples given that the side lengths are [tex]\( x^2 - 1 \)[/tex], [tex]\( 2x \)[/tex], and [tex]\( x^2 + 1 \)[/tex], for [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex], follow these steps:
1. For [tex]\( x = 3 \)[/tex]
- Calculate the first side length using [tex]\( x^2 - 1 \)[/tex]:
[tex]\[
3^2 - 1 = 9 - 1 = 8
\][/tex]
- Calculate the second side length using [tex]\( 2x \)[/tex]:
[tex]\[
2 \cdot 3 = 6
\][/tex]
- Calculate the hypotenuse using [tex]\( x^2 + 1 \)[/tex]:
[tex]\[
3^2 + 1 = 9 + 1 = 10
\][/tex]
- Arrange the sides in ascending order:
[tex]\[
(6, 8, 10)
\][/tex]
2. For [tex]\( x = 5 \)[/tex]
- Calculate the first side length using [tex]\( x^2 - 1 \)[/tex]:
[tex]\[
5^2 - 1 = 25 - 1 = 24
\][/tex]
- Calculate the second side length using [tex]\( 2x \)[/tex]:
[tex]\[
2 \cdot 5 = 10
\][/tex]
- Calculate the hypotenuse using [tex]\( x^2 + 1 \)[/tex]:
[tex]\[
5^2 + 1 = 25 + 1 = 26
\][/tex]
- Arrange the sides in ascending order:
[tex]\[
(10, 24, 26)
\][/tex]
With this, we have the following table with the missing values filled in:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex]-value & Pythagorean Triple \\
\hline
3 & [tex]$(6,8,10)$[/tex] \\
\hline
5 & [tex]$(10,24,26)$[/tex] \\
\hline
\end{tabular}
So the correct triples are [tex]\( (6,8,10) \)[/tex] for [tex]\( x = 3 \)[/tex] and [tex]\( (10,24,26) \)[/tex] for [tex]\( x = 5 \)[/tex].
