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Lesson 4.5 , For use with pages 266-271

1. √49. 2. –√144. Lesson 4.5 , For use with pages 266-271. Find the exact value. 7. ANSWER. –12. ANSWER. 82. 3. Use calculator to approximate the value of to the nearest tenth. 16. 1.

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Lesson 4.5 , For use with pages 266-271

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  1. 1. √49 2. –√144 Lesson 4.5, For use with pages 266-271 Find the exact value. 7 ANSWER –12 ANSWER

  2. 82 3. Use calculator to approximate the value of to the nearest tenth. 16 1 4. The area of half of a square mural is 60 square feet. What is the length of a side of the mural? 2 Lesson 4.5, For use with pages 266-271 ANSWER 2.3 ANSWER 11 ft

  3. 4.5 Properties of Square Roots

  4. = 4 = a. = = b. 7 = 3 5 9 4 4 5 7 7 6 2 9 4 16 80 14 81 14 16 21 = c. 81 16 = 126 = d. = EXAMPLE 1 Use properties of square roots Simplify the expression.

  5. = 3 = 7 = 3 3 2 2 98 98 9 27 27 49 = for Example 1 GUIDED PRACTICE GUIDED PRACTICE SOLUTION SOLUTION

  6. 10 8 10 8 = = 5 = 6 14 6 25 15 15 28 28 16 224 150 = = 14 = 4 for Example 1 GUIDED PRACTICE GUIDED PRACTICE SOLUTION SOLUTION

  7. 9 15 9 9 15 = 15 3 8 4 64 = = 64 64 4 4 15 = 2 for Example 1 GUIDED PRACTICE GUIDED PRACTICE SOLUTION SOLUTION

  8. 11 36 36 11 11 11 = = 25 49 5 25 49 49 25 36 6 = = 7 for Example 1 GUIDED PRACTICE GUIDED PRACTICE SOLUTION SOLUTION

  9. 3 7 + 5 2 2 2 5 5 2 2 5 (a) = 2 2 = 10 = 2 EXAMPLE 2 Rationalize denominators of fractions. Simplify and (a) (b) SOLUTION

  10. 3 3 7 + 7 + 7 – (b) = 7 – 2 2 2 2 2 2 2 2 21 – 3 = 49 – 7 + 7 – 2 21 – 3 = 47 EXAMPLE 2 Rationalize denominators of fractions. SOLUTION

  11. 3 3 4 + x = 12 + x = + 2 x = EXAMPLE 3 Solve a quadratic equation Solve 3x2 + 5 = 41. 3x2 + 5 = 41 Write original equation. 3x2 = 36 Subtract 5 from each side. x2 = 12 Divide each side by 3. Take square roots of each side. Product property Simplify.

  12. – 2 2 3 3 3 3( )2 + 5 = 41 ? 3( )2 + 5 = 41 ? 2 – 2 3 3(12) + 5 = 41 ? 3(12) + 5 = 41 ? 41 = 41 41 = 41   EXAMPLE 3 Solve a quadratic equation ANSWER The solutions are and Check the solutions by substituting them into the original equation. 3x2 + 5 = 41 3x2 + 5 = 41

  13. 35 35 35 35 (z + 3)2 = 7 15 z + 3 = + z = – 3 + The solutions are – 3 + and – 3 – EXAMPLE 4 Standardized Test Practice SOLUTION Write original equation. (z + 3)2 = 35 Multiply each side by 5. Take square roots of each side. Subtract 3 from each side.

  14. EXAMPLE 4 Standardized Test Practice ANSWER The correct answer is C.

  15. 5 5 5 6 6 6 6 5 = 5 5 = 30 5 = for Examples 2, 3, and 4 GUIDED PRACTICE GUIDED PRACTICE Simplify the expression. SOLUTION

  16. 9 8 8 9 9 8 8 9 = 8 8 = 3 2 = 4 for Examples 2, 3, and 4 GUIDED PRACTICE GUIDED PRACTICE Simplify the expression. SOLUTION

  17. 17 17 17 12 12 12 12 17 = 12 12 = 2 51 51 = = 12 6 for Examples 2, 3, and 4 GUIDED PRACTICE GUIDED PRACTICE Simplify the expression. SOLUTION

  18. 19 21 21 21 19 19 19 21 = 21 21 = 399 = 21 for Examples 2, 3, and 4 GUIDED PRACTICE GUIDED PRACTICE Simplify the expression. SOLUTION

  19. 7 + 7 + 5 5 5 5 5 5 5 5 5 – 6 – 6 – 6 = 7 – 7 – 7 – – 42 – 6 = 49 – 7 + 7 – 5 – 21 – 3 = 22 for Examples 2, 3, and 4 GUIDED PRACTICE SOLUTION

  20. 11 11 11 11 11 11 11 11 11 4 – 2 2 2 = 4 + 4 + 4 + 4 – 8 – 2 = 16 – 4 + 4 – 11 8 – 2 = 5 for Examples 2, 3, and 4 GUIDED PRACTICE SOLUTION

  21. 7 7 7 7 7 7 7 7 7 9 – – 1 – 1 – 1 = 9 + 9 + 9 + 9 – – 9 + = 81 – 9 + 9 – 7 – 9 + = 74 for Examples 2, 3, and 4 GUIDED PRACTICE SOLUTION

  22. 3 3 3 3 3 3 3 3 3 8 + 4 4 4 = 8 – 8 – 8 – 8 + 32 + 4 = 64 – 4 + 4 – 3 32 + 4 = 61 for Examples 2, 3, and 4 GUIDED PRACTICE SOLUTION

  23. 4 4 + x = 16 + x = + 4 x = for Examples 2, 3, and 4 GUIDED PRACTICE Solve the equation. 5x2 = 80 SOLUTION 5x2 = 80 Write original equation. x2 = 16 Divide each side by 5. Take square roots of each side. Product property Simplify.

  24. ? 5(4)2 = 80 ? 5(– 4)2 = 80 ? ? 5(16) = 80 5(16) = 80 80 = 80 80 = 80   for Examples 2, 3, and 4 GUIDED PRACTICE ANSWER The solutions are and . 4 – 4 CheckCheck the solutions by substituting them into the original equation. 5x2 = 80 5x2 = 80

  25. 6 6 + z = 36 + z = + 6 z = for Examples 2, 3, and 4 GUIDED PRACTICE Solve the equation. z2 – 7 = 29 SOLUTION z2 – 7 = 29 Write original equation. Add 7 to each side. z2 = 36 Take square roots of each side. Product property Simplify.

  26. (– 6)2 – 7 = 29 ? (6)2 – 7 = 29 ? ? 36– 7 = 29 36– 7 = 29 ? 29 = 29 29 = 29   for Examples 2, 3, and 4 GUIDED PRACTICE ANSWER The solutions are and . 6 – 6 Checkthe solutions by substituting them into the original equation. z2 – 7 = 29 z2 – 7 = 29

  27. 3 3 40 (x – 2)2 = 3 40 + (x – 2) = 3 40 40 + + x = x = 2 2 3 3 120 + = 2 3 for Examples 2, 3, and 4 GUIDED PRACTICE 3(x – 2)2 = 40 SOLUTION 3(x – 2)2 = 40 Write original equation. Divide each side by 3. Take square roots of each side. Division property

  28. 2 30 2 30 + + = 2 2 3 3 ANSWER 4(30) The solutions are . + = 2 3 for Examples 2, 3, and 4 GUIDED PRACTICE

  29. For a science competition, students must design a container that prevents an egg from breaking when dropped from a height of 50 feet. How long does the container take to hit the ground ? EXAMPLE 5 Model a dropped object with a quadratic function Science Competition

  30. 50 = t2 5016 16 + = t2 + 1.8 t EXAMPLE 5 Model a dropped object with a quadratic function SOLUTION h = – 16t 2+ h0 Write height function. Substitute 0 for hand 50 for h0. 0 = – 16t 2 + 50 Subtract 50 from each side. – 50 = – 16t 2 Divide each side by – 16. Take square roots of each side. Use a calculator.

  31. EXAMPLE 5 Model a dropped object with a quadratic function ANSWER Reject the negative solution, – 1.8 , because time must be positive. The container will fall for about 1.8 seconds before it hits the ground.

  32. 30 3016 16 + = t2 + 1.4 t = t2 for Example 5 GUIDED PRACTICE GUIDED PRACTICE What If?In Example 5, suppose the egg container is dropped from a height of 30feet. How long does the container take to hit the ground? SOLUTION h = – 16t 2+ h0 Write height function. Substitute 0 for hand 30 for h0. 0 = – 16t 2 + 30 Subtract 30 from each side. – 30 = – 16t 2 Divide each side by – 16. Take square roots of each side. Use a calculator.

  33. for Example 5 GUIDED PRACTICE GUIDED PRACTICE ANSWER Reject the negative solution, – 1.4, because time must be positive. The container will fall for about 1.4 seconds before it hits the ground.

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