Presentation is loading. Please wait.

Presentation is loading. Please wait.

Core Focus on Geometry Exponents and Roots Lesson 4.5.

Similar presentations


Presentation on theme: "Core Focus on Geometry Exponents and Roots Lesson 4.5."— Presentation transcript:

1 Core Focus on Geometry Exponents and Roots Lesson 4.5

2 Warm-Up Solve each equation. 1. 2. 3. x = 9 x = −4 x = 100

3 Exponents and Roots Lesson 4.5 Use roots to solve equations with exponents.

4 Vocabulary Square Root One of the two equal factors of a number. Symbol is. Perfect Square A number whose square root is an integer. Cube Root One of the three equal factors of a number. Symbol is. Perfect Cube A number whose cube root is an integer. The First 10 Perfect Cubes

5 When finding the root of a number, the answer may be positive, negative or both. For example, if x 2 = 9, then x can either equal 3 or –3 because (3)(3) = 9 and (−3)(−3) = 9. When solving problems involving square and cube roots, consider the following rules:  Terms with exponents that are even will have two root answers, positive and negative. This is shown with the symbol ±.  Terms with exponents that are odd will have one root answer. It will have the same sign as the term under the root.  In real-world situations, negative answers may not always make sense. Look closely at the situation to determine if only a positive answer is needed. Good to Know!

6 Using Roots to Solve Equations with Exponents 1. Isolate the variable with the exponent. 2. Use inverse operations to undo the exponent with the corresponding root. 3. Determine if the answer should be positive, negative or both based on the original exponent and the application setting.

7 Example 1 Solve for x. Include all answers. Subtract 11 from both sides of the equation. Multiply both sides of the equation by 2. Square root both sides of the equation. Include both positive and negative answers. (–8)(–8) = 64 and (8)(8) = 64

8 Example 2 Solve 2x 3 − 5 = −59 for x. Add 5 to both sides of the equation. Divide both sides of the equation by 2. Cube root both sides of the equation. (–3)(–3)(–3) = –27

9 Example 3 The volume of a sphere can be found using the formula. Finley has a spherical balloon filled with water. She knows there are 56 cubic inches of water in the balloon. What is the approximate radius, r, of the balloon? Use 3.14 for π and round the answer to the nearest hundredth. Write the formula for the volume of a sphere. Substitute all known values for the variables. Multiply both sides of the equation by the reciprocal of.

10 Example 3 Continued… The volume of a sphere can be found using the formula. Finley has a spherical balloon filled with water. She knows there are 56 cubic inches of water in the balloon. What is the approximate radius, r, of the balloon? Use 3.14 for π and round the answer to the nearest hundredth. Divide both sides of the equation by 3.14. Cube root both sides of the equation. The radius of the balloon is approximately 2.37 inches.

11 Some equations in this lesson have one solution and others have two solutions. Is it possible to tell how many solutions an equation will have prior to solving? If so, explain how. Communication Prompt

12 Exit Problems Solve each equation. Include all answers. 1.2. 3.4. x = 7 x = ±4 x = ±10 x = −5


Download ppt "Core Focus on Geometry Exponents and Roots Lesson 4.5."

Similar presentations


Ads by Google