Question

What is the form of the Two Squares identity?+O A. (2 + b)(c2 + ) = (ac- boy? +(ad + bc)?B. (22 - 6?)(c + 2) = (ac- borje - (ad + bc)?c. (82 +62)(c2 - 02) - (ac+ bc)? - (ad + bc)?D. (22 +62)(c2 + 2) - (2b-cdj ° + (ac + bc)

Answer 1

For this exercise, let's check every option given in the picture:

Option A

Given:

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

Simplify it by solving the multiplication:

$\begin{gathered} =(a^2)(c^2)+(a^2)(d^2)+(b^2)(c^2)+(b^2)(d^2) \\ =a^2c^2+a^2d^2+b^2c^2+b^2d^2 \end{gathered}$

Notice that the expression given in the exercise as the product is:

$(ac-bd)^2+(ad+bc)^2$

Îf you simplify it, you get:

$\begin{gathered} =(ac)^2-2(ac)(bd)+(bd)^2+(ad)^2+2(ad)(bc)+(bc)^2 \\ =a^2c^2+b^2d^2+a^2d^2+b^2c^2 \end{gathered}$

Therefore:

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

Option B

Apply the same steps used in Option A. Then:

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

Simplify the result given in the exercise :

$\begin{gathered} (ac-bd)^2-(ad+bc)^2=(ac)^2-2(ac)(bd)+(bd)^2-((ad)^2+2(ad)(bc)+(bc)^2) \\ =a^2c^2-a^2d^2-4abcd-b^2c^2+b^2d^2 \\ \end{gathered}$

Compare it with the one found above:

$a^2c^2+a^2d^2-b^2c^2-b^2d^2\ne a^2c^2-2(ac)(bd)+b^2d^2-a^2d^2-2(ad)(bc)-b^2c^2$

Option C

Applying the same procedure, you get:

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

Simplify the result shown in the picture:

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

Notice that:

$a^2c^2-a^2d^2+b^2c^2-b^2d^2\ne a^2c^2+2(ac)(bd)+b^2d^2-a^2d^2-2(ad)(bc)-b^2c^2$

Option D

You already know that:

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

Simplify the result given in the exercise:

$(ab-cd)^2+(ac+bd)^2=a^2b^2+a^2c^2+b^2d^2+c^2d^2$

Then:

$(a^2+b^2)(c^2+d^2)\ne(ab-cd)^2+(ac+bd)^2$

The answer is: OptionA.