Let's solve Part A by proving that the triangle with side lengths [tex]\(a = x^2 - 1\)[/tex], [tex]\(b = 2x\)[/tex], and [tex]\(c = x^2 + 1\)[/tex] is a right triangle when [tex]\(x > 1\)[/tex]. We will use the Pythagorean theorem for this purpose.
The Pythagorean theorem states that for a right triangle with side lengths [tex]\(a\)[/tex], [tex]\(b\)[/tex], and hypotenuse [tex]\(c\)[/tex], the following equation holds:
[tex]\[a^2 + b^2 = c^2.\][/tex]
Using the given side lengths, let's create and simplify this equation step-by-step.
Step 1: Write down the square of each side.
[tex]\[ a^2 = (x^2 - 1)^2 \][/tex]
[tex]\[ b^2 = (2x)^2 \][/tex]
[tex]\[ c^2 = (x^2 + 1)^2 \][/tex]
Step 2: Expand these expressions.
First, expand [tex]\(a^2 = (x^2 - 1)^2\)[/tex]:
[tex]\[ a^2 = (x^2 - 1)(x^2 - 1) \][/tex]
[tex]\[ a^2 = x^4 - 2x^2 + 1 \][/tex]
Next, expand [tex]\(b^2 = (2x)^2\)[/tex]:
[tex]\[ b^2 = 4x^2 \][/tex]
Finally, expand [tex]\(c^2 = (x^2 + 1)^2\)[/tex]:
[tex]\[ c^2 = (x^2 + 1)(x^2 + 1) \][/tex]
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
Step 3: Add [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] to see if it equals [tex]\(c^2\)[/tex].
[tex]\[ a^2 + b^2 = (x^4 - 2x^2 + 1) + 4x^2 \][/tex]
[tex]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 \][/tex]
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
Step 4: Compare [tex]\(a^2 + b^2\)[/tex] with [tex]\(c^2\)[/tex].
From previous steps:
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
We observe that:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Since the left-hand side of the equation equals the right-hand side of the equation, the given side lengths [tex]\(a = x^2 - 1\)[/tex], [tex]\(b = 2x\)[/tex], and [tex]\(c = x^2 + 1\)[/tex] satisfy the Pythagorean theorem. Hence, the given triangle is a right triangle when [tex]\(x > 1\)[/tex].
