Question

Given the following two pair for values x1 = 5, y1 = 4 and x 2 = 1, y2 = .
Calculate the Pythagorean triples associated with each pair.
Based on the results of the previous activity, for which values xn and yn is this identity valid?

Answer 1

Answer:

Given the following two pair for values x 1 = 5, y 1 = 4 and x 2 = 1, y 2 = .

Calculate the Pythagorean triples associated with each pair

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x 1 = 5, y 1 = 4

Ans:: sqrt[5^2+4^2] = sqrt[41

 

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and x 2 = 1, y 2 = ?

Ans: r = sqrt[1^2 + ?^2]

Answer 2

Answer:

Part 1: 1681=1681 and 1.5625=1.5625

Part 2: Both Identities are valid

Step-by-step explanation:

PART 1

Pathagorean Triples Formula used

$(x^{2} +y^{2} )^{2} =(x^{2} -y^{2} )^{2} +(2xy)^{2}$

First you input the first pair which is x=5 and y=4

$(5^{2} +4^{2} )^{2} =(5^{2} -4^{2} )^{2} +(2(5)(4))^{2}$

Then, solve for each part of the equation

1681 = 81 + 1600

1681 = 1681

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Now do the same for the next pair x=1 and y=$\frac{1}{2}$

$(1^{2} +\frac{1}{2} ^{2} )^{2} =(1^{2} -\frac{1}{2}^{2} )^{2} +(2(1)(\frac{1}{2} ))^{2}$

Then, solve for each part of the equation

1.5625 = 0.5625 + 1

1.5625 = 1.5625

Part 2

Both identities are valid.