To identify the Pythagorean triples for given [tex]\(x\)[/tex]-values and to find a missing [tex]\(x\)[/tex]-value that completes a triple, let's go through the steps methodically:
Given side lengths for a right triangle:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ b = 2x \][/tex]
[tex]\[ c = x^2 + 1 \][/tex]
We know these side lengths must satisfy the Pythagorean theorem:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Let's validate this:
[tex]\[ (x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2 \][/tex]
[tex]\[ x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 \][/tex]
[tex]\[ x^4 + 2x^2 + 1 = x^4 + 2x^2 + 1 \][/tex]
The equation verifies the side lengths.
Now, let's find the specific triples:
1. When [tex]\( x = 3 \)[/tex]:
[tex]\[ a = 3^2 - 1 = 9 - 1 = 8 \][/tex]
[tex]\[ b = 2 \times 3 = 6 \][/tex]
[tex]\[ c = 3^2 + 1 = 9 + 1 = 10 \][/tex]
Hence, the Pythagorean triple is [tex]\((8,6,10)\)[/tex].
2. When [tex]\( x = 5 \)[/tex]:
Already given:
[tex]\[ a = 5^2 - 1 = 25 - 1 = 24 \][/tex]
[tex]\[ b = 2 \times 5 = 10 \][/tex]
[tex]\[ c = 5^2 + 1 = 25 + 1 = 26 \][/tex]
Hence, the Pythagorean triple is [tex]\((24,10,26)\)[/tex].
3. Another valid [tex]\( x \)[/tex]-value must be greater than 5, let's choose [tex]\( x = 6 \)[/tex]:
[tex]\[ a = 6^2 - 1 = 36 - 1 = 35 \][/tex]
[tex]\[ b = 2 \times 6 = 12 \][/tex]
[tex]\[ c = 6^2 + 1 = 36 + 1 = 37 \][/tex]
Hence, the new Pythagorean triple is [tex]\((35,12,37)\)[/tex].
Thus, the completed table with the correct answers is:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex]-value & Pythagorean Triple \\
\hline
3 & (6,8,10) \\
5 & (8,15,17) \\
6 & (12,35,37) \\
\hline
\end{tabular}
