1. Algebra 1
Section 11.3

2. Product Property of Radicals
For x ≥ 0 and y ≥ 0,
x • y = xy n

3. Example 1
Simplify 2 •
= 2(2 • 3)
= • 3
= 2 3

4. Example 2 Simplify -2 18 • 5 50 . = -2 • 5 18 • 50
= -2 • • 50
= (2 • 32)(2 • 52)
= -10 • 2 • 3 • 5
= -300

5. Multiplying Radicals
From here on we will assume that all variables in the radicand represent nonnegative values.
Therefore, it will not be necessary to use absolute values.

6. Example 3
Simplify 6ab • 2bc .
= (3 • 2)2ab2c
= 2b 3ac

7. Multiplying Radicals
Apply the Product Property, writing the radicand as the product of its prime factors.
Simplify the root of each perfect power.
Simplify the resulting expression.

8. Example 4 Simplify 225 • 315 . = (32 • 52)(32 • 5 • 7)
= (32 • 52)(32 • 5 • 7) 3
= • 3 • 53 • 7 3
= 3 • • 7 3 = 3

9. Multiplying Radicals
More complicated expressions involving radicals can also be simplified using the Product Property by applying other previously established principles.

10. Example 7
a) Simplify 5 ( 3 – 2).
= – 2 5

11. Example 7 b) Simplify 2 6 ( 15 – 3 10). = 2 (2 • 3)(3 • 5)
= 2 (2 • 3)(3 • 5)
– 6 (2 • 3)(2 • 5)
= 2 • • 5 – 6 • • 5
= –

12. Heron’s Formula
To calculate the area of a triangle from the lengths of its three sides:
A = s(s – a)(s – b)(s – c)
where s =
a + b + c 2

13. Example 8 Sides: 6 ft, 8 ft, and 10 ft. s = 6 + 8 + 10 2 = 12 ft

14. Homework: pp