2. Algebra I Concept Test # 12 – Square Roots 1. 75 5 15 Product Property of Radicals 5 3 25 3 Simplify: 2. 81x 10 Product Property of Radicals 81 x 10 9x 5 © 2007- 10 by S-Squared, Inc. All Rights Reserved.

3. 3 Algebra I Concept Test # 12 – Square Roots 4 Simplify: 2 1 3. 64 Simplify 2 1 8 4 4. 32 18 Reduce 16 9 Quotient Property of Radicals 16 9 Simplify

4. Algebra I Concept Test # 12 – Square Roots 6 x 2 y 10 5. 6x 2 y 10 Product Property of Radicals x y 5 6 Simplify:

5. Rewrite Radicand with perfect square factor ─ 6 7 ─ 3 2 7 2 7 ─ 3 28 2 7 Term is in simplest form, Radicand is prime ─ 3 4 7 2 7 Square Root 2 7 − 4 7 Multiply Subtract Simplify: 6. Algebra I Concept Test # 12 – Square Roots

6. 7. Approximate the following square roots to the tenths place: a)10 **Notice the square root of 9 is 3 and the square root of 16 is 4. Since 10 is between 9 and 16, the square root of 10 must be between 3 and 4. Approximately 3.2 b)79 **Notice the square root of 64 is 8 and the square root of 81 is 9. Since 79 is between 64 and 81, the square root of 79 must be between 8 and 9. Approximately 8.9

7. Algebra I Concept Test # 12 – Square Roots 8.Evaluate: a + bc for a = 8, b = − 4, and c = − 9 2 Substitute (8) 2 + (−4)(−9) Exponent 64 + (−4)(−9) 64 + 36 100 10 Multiply Add Square Root

8. ( 3 ─ 8 ) 5 15 9.Distribute ─ 8 5 Algebra I Concept Test # 12 – Square Roots 64 a 2 b 4 Simplify: 10. 8a 3 b 8a −1 b 3 8ab2 8ab2 Multiply Product Property of radicals Square Root

9. ( ) 2 3 x = 15 7 + 3 x = 22 Solve: 11. Isolate the radical x Subtract Divide Square 3 3 – 7 x = 5 ( ) 2 x = 25 Algebra I Concept Test # 12 – Square Roots

10. ( ) 2 x = 7 ( ) 2 6 x = 42 2 x + 4 x = 42 Solve: 12. Combine like terms Divide Square 6 6 x = 49 13.Check your solution for #12. 2 x + 4 x = 42 Substitute Simplify Check Complete 2 49 + 4 49 = 42 2 7 + 4 7 = 42 14 + 28 = 42 42 = 42 Algebra I Concept Test # 12 – Square Roots

11. x = − 4 − 6 x = 24 5 x ─ 11 x ─ 10 = 14 Solve: 14. Isolate the radical x Add Divide − 6 + 10 Combine like terms − 6 x ─ 10 = 14 Note: The square root of a number can never be negative No Solution Algebra I Concept Test # 12 – Square Roots

12. Note: Substitute each of the given x-coordinates into the given radical equation to find the y-coordinate. 15.Using the square root function, y = x – 7, complete the following: y = x – 7 xy 1/4 − 6 1/2 0 1 4 − 6 − 7 − 5 a)Complete the table b)Graph the function using the ordered pairs from part a. 1 2 3 − 1 − 5 − 4 − 3 − 2 x y − 5 − 1 − 2 − 3 − 4 5 4 3 2 1 − 7 − 6

13. Algebra I Concept Test # 12 – Square Roots (49, 0) 15.Using the square root function, y = x – 7, complete the following: c)Identify the endpoint d)Identify the y-intercept (0, − 7) e)Identify the x-intercept (0, − 7) g)State the range y ≥ − 7 f)State the domain x ≥ 0 1 2 3 − 1 − 5 − 4 − 3 − 2 x y − 5 − 1 − 2 − 3 − 4 5 4 3 2 1 − 7 − 6 xy 1/4 − 6 1/2 0 1 4 − 6 − 7 − 5 Note: Let y = 0 and then solve for x. 0 = x – 7 7 = x 49 = x RADICAL