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Relating a Polynomial Identity to Pythagorean Triples

In this activity, you will relate polynomial identities to Pythagorean triples. Answer the following question based on a triangle with side lengths [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex].

Part A
Prove that when [tex]\( x \ \textgreater \ 1 \)[/tex], a 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. Use the Pythagorean theorem and the given side lengths to create an equation. Use the equation to show that this triangle follows the rule describing right triangles. Explain your steps.

Sagot :

GinnyAnswer
Alright, let's go through the process of proving that a 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'll use the Pythagorean theorem to do this.

### Step-by-Step Solution:

1. State the Pythagorean theorem:
The Pythagorean theorem states that in a right triangle with legs [tex]\( a \)[/tex] and [tex]\( b \)[/tex], and hypotenuse [tex]\( c \)[/tex], the following equation holds:
[tex]\[ a^2 + b^2 = c^2 \][/tex]

2. Insert given expressions for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
The side lengths given are [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex]. Let's compute [tex]\( a^2 \)[/tex], [tex]\( b^2 \)[/tex], and [tex]\( c^2 \)[/tex].

3. Calculate [tex]\( a^2 \)[/tex]:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ a^2 = (x^2 - 1)^2 \][/tex]
Expanding the square:
[tex]\[ a^2 = (x^2 - 1)(x^2 - 1) = x^4 - 2x^2 + 1 \][/tex]

4. Calculate [tex]\( b^2 \)[/tex]:
[tex]\[ b = 2x \][/tex]
[tex]\[ b^2 = (2x)^2 = 4x^2 \][/tex]

5. Calculate [tex]\( c^2 \)[/tex]:
[tex]\[ c = x^2 + 1 \][/tex]
[tex]\[ c^2 = (x^2 + 1)^2 \][/tex]
Expanding the square:
[tex]\[ c^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1 \][/tex]

6. Compare [tex]\( a^2 + b^2 \)[/tex] with [tex]\( c^2 \)[/tex]:
Now, let's add [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[ a^2 + b^2 = (x^4 - 2x^2 + 1) + 4x^2 \][/tex]
Simplify the expression:
[tex]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 \][/tex]

We have computed [tex]\( c^2 \)[/tex] as:
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]

Notice that:
[tex]\[ a^2 + b^2 = c^2 \][/tex]

7. Conclusion:
Since [tex]\( a^2 + b^2 = c^2 \)[/tex] holds true, we have shown that the triangle with sides [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] satisfies the Pythagorean theorem.

Therefore, it is proven that for [tex]\( x > 1 \)[/tex], a triangle with side lengths [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] is indeed a right triangle.
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