1. EXAMPLE 6 Multiply radical expressions a. Simplify. ( 4 – 5 Product property of radicals 20 Distributive property Simplify. ) = 4–2055 = 45– 100 = – 1045 b. + –3 ( )() 2727 = 7 273 – + 2 – 3 () 2 = 14 7 – 3 + – 6 = 1 – 2 Multiply. Product property of radicals Simplify. = ( ) ( ) ( + 2+ + –3 2 7727–3 ) 22

2. EXAMPLE 7 Solve a real-world problem ASTRONOMY a. Simplify the formula. b. Jupiter’s average distance from the sun is shown in the diagram. What is Jupiter’s orbital period? The orbital period of a planet is the time that it takes the planet to travel around the sun. You can find the orbital period P (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

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

4. GUIDED PRACTICE for Examples 6 and 7 Simplify the expression 7. ( ) 5 4 – ( ) 1 – 5 = 9 – 55 8. Neptune’s average distance from the sun is about 6 t imes Jupiter’s average distance from the sun. Is the orbital period of Neptune 6 times the orbital period of Jupiter? Explain. ASTRONOMY 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.