BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.3, Problem 131E
To determine

To show: The algebraic identity (a2+b2)(c2+d2)=(ac+bd)2+(adbc)2 holds.

Expert Solution

Explanation of Solution

The given algebraic identity is (a2+b2)(c2+d2)=(ac+bd)2+(adbc)2.

Solve the expression (ac+bd)2+(adbc)2.

(ac+bd)2+(adbc)2=(a2c2+b2d2+2abcd)+(a2d2+b2c22abcd)=(a2c2+b2d2+a2d2+b2c2)=(a2c2+a2d2+b2d2+b2c2)

Note that, the common factor in a2c2+a2d2 is a2 and in b2d2+b2c2 is b2. That is,

(a2c2+a2d2+b2d2+b2c2)=a2(c2+d2)+b2(c2+d2)=(a2+b2)(c2+d2)

Thus, the algebraic identity of the left side and right side is equal.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!