1. 4 – = 14 14 14 20 20 4 – 100 = 4 – 10 = ( ) ( ) b. – 3 + ( ( ) ) ) ( 2 – 3 = + – 3 + + ( ) 2 – – 7 3 3 = + – – 3 6 = 7 + 7 2 7 2 2 7 7 7 2 2 2 2 7 2 7 2 – 1 2 = 5 5 5 5 5 EXAMPLE 6 Multiply radical expressions (4 – ) Distributive property a. Product property of radicals Simplify. Multiply. Product property of radicals Simplify. Simplify.

2. The orbital period of a planet is the time that it takes the planet to travel around the sun. You can find the orbital periodP(in Earth years) using the formula P = d where d is the average distance (in astronomical units, abbreviated AU) of the planet from the sun. 3 Simplify the formula. a. Jupiter’s average distance from the sun is shown in the diagram. What is Jupiter’s orbital period? b. EXAMPLE 7 Solve a real-world problem ASTRONOMY

3. 5.2 5.2 a. P d = 3 dd = 2 d d = b. Substitute 5.2 for din the simplified formula. 5.2 P d d = = ANSWER The orbital period of Jupiter is 5.2 , or about 11.9, Earth years. EXAMPLE 7 Solve a real-world problem SOLUTION Write formula. Factor using perfect square factor. d d Product property of radicals = 2 Simplify.

4. ) ) ( ( 1 – 4 – – 9 5 5 = ASTRONOMY 8. Neptune’s average distance from the sun is about 6 times Jupiter’s average distance from the sun. Is the orbital period of Neptune 6 times the orbital period of Jupiter? Explain. 5 5 No. Neptune’s orbital period is (6 5) (6 5) = 6 5 6 5 = 6 6 5 5. Thus, Neptune’s orbital period is 6 6 times the orbital period of Jupiter. for Examples 6 and 7 GUIDED PRACTICE 7. Simplify the expression