Question

which identity is the result of using the pythagorean theorem to show that a triangle with side lengths ( x^2 - 1 ), ( 2x ), and ( x^2 + 1 ) is a right triangle? select the correct answer. ( circ - (x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2 ) ( circ (x^2 - 1)^2 - (2x)^2 = (x^2 + 1)^2 ) ( circ (x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2 ) ( circ (x^2 - 1)^2 + (2x)^2 = - (x^2 + 1)^2 )

Explanation:

Step1: Recall Pythagorean Theorem

The Pythagorean theorem states that for a right triangle with legs \(a\), \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\), where the hypotenuse is the longest side. First, we need to determine which side is the hypotenuse among \(x^{2}-1\), \(2x\) and \(x^{2}+1\). Since \(x^{2}+1\) is larger than \(x^{2}-1\) (because \(1>- 1\)) and for \(x
eq0\), \(x^{2}+1>2x\) (since \(x^{2}+1 - 2x=(x - 1)^{2}\geq0\), and equality holds when \(x = 1\)), so \(x^{2}+1\) is the hypotenuse. The legs are \(x^{2}-1\) and \(2x\).

Step2: Apply Pythagorean Theorem

According to the Pythagorean theorem, \(a^{2}+b^{2}=c^{2}\), where \(a=x^{2}-1\), \(b = 2x\) and \(c=x^{2}+1\). Substituting these values into the formula, we get \((x^{2}-1)^{2}+(2x)^{2}=(x^{2}+1)^{2}\).

Answer:

\((x^{2}-1)^{2}+(2x)^{2}=(x^{2}+1)^{2}\) (the third option from the top, i.e., the option \((x^{2}-1)^{2}+(2x)^{2}=(x^{2}+1)^{2}\))