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What is the form of the Two Squares identity?
O A. (a? + b?)(C2 - 082) - (ac+ bd)? - (ad + bc)
O B. (a + b)(c? + 2) = (ac- boy2 + ( ad + bc)?
O C. (22 +62) (c2 + (x2) - (ab- cc)* + (ac + boy?
O D. (a? - 63) (c? + 2) = (ac- boj? - (ad + bc)?
SUBMIT

Respuesta :

The two squares identity is (c) (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2

How to determine the identity?

The complete question is in the attached image

The identity is given as:

Two Squares identity

The Two Squares identity states that:

The difference of two squares is the product of the sum of the square roots and their difference

This means that

(a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2

Rewrite as:

(a^2 + b^2)(c^2 + d^2) = a^2c^2  + b^2d^2 + a^2d^2 + b^2c^2

Add 0 to the equation

(a^2 + b^2)(c^2 + d^2) = a^2c^2  + b^2d^2 + a^2d^2 + b^2c^2 + 0

Express 0 as 2abcd - 2abcd

(a^2 + b^2)(c^2 + d^2) = a^2c^2  + b^2d^2 + a^2d^2 + b^2c^2 + 2abcd - 2abcd

Rewrite as:

(a^2 + b^2)(c^2 + d^2) = a^2c^2 + b^2d^2 - 2abcd + a^2d^2 + b^2c^2 + 2abcd

Group the terms

(a^2 + b^2)(c^2 + d^2) = (a^2c^2 + b^2d^2 - 2abcd) + (a^2d^2 + b^2c^2 + 2abcd)

Factorize each group

(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2

Hence, the two squares identity is (c) (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2

Read more about two squares identity at:

https://brainly.com/question/4656865

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Ver imagen MrRoyal