To determine which identity shows that a triangle with side lengths [tex]\( x^2 - 1 \)[/tex], [tex]\( 2x \)[/tex], and [tex]\( x^2 + 1 \)[/tex] is a right triangle, we will use the Pythagorean theorem.
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. If we have a triangle with sides [tex]\( a \)[/tex], [tex]\( b \)[/tex], and hypotenuse [tex]\( c \)[/tex], the equation is given by:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
We need to verify if the following side lengths satisfy this equation:
- [tex]\( a = x^2 - 1 \)[/tex]
- [tex]\( b = 2x \)[/tex]
- [tex]\( c = x^2 + 1 \)[/tex]
Let's square each of these terms and see if they satisfy the Pythagorean theorem.
1. Square the first side:
[tex]\[ (x^2 - 1)^2 = x^4 - 2x^2 + 1 \][/tex]
2. Square the second side:
[tex]\[ (2x)^2 = 4x^2 \][/tex]
3. Add the squares of the first and second sides:
[tex]\[ (x^2 - 1)^2 + (2x)^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 \][/tex]
4. Square the hypotenuse:
[tex]\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \][/tex]
We observe that:
[tex]\[ (x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2 \][/tex]
[tex]\[ x^4 + 2x^2 + 1 = x^4 + 2x^2 + 1 \][/tex]
This matches the identity:
[tex]\[ (x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{\left(x^2-1\right)^2+(2 x)^2=\left(x^2+1\right)^2} \][/tex]
