Solving Radical Equations: Key Concepts and Common Errors

1. Introduction Using the Pythagorean Theorem to solve problems provides a familiar example of a relationship between variables that involve radicals (or roots). Recall that the word root in mathematics has several definitions. You have seen and worked with function roots, or zeros; in this lesson, you will work with a different kind of root—that is, square roots, cube roots, and other radicals. 3.2.3: Solving Radical Equations

2. Introduction, continued Radical expressions, or expressions containing a root, follow rules that may seem more complicated than those of more common numbers. While you can easily add the numbers 4 and 9 to get 13, there is no similar simple way to add the radicals and without simplifying each individual term first. 3.2.3: Solving Radical Equations

3. Introduction, continued Since radical expressions require more complex methods in order to be simplified, radical equations also involve more complex methods in order to be solved. However, some common real-life situations can be more readily solved by applying radical equations, making fluency with radical equations a useful skill to learn. In this lesson, we will explore ways to solve radical equations, as well as how to assess and validate the resulting solutions. 3.2.3: Solving Radical Equations

4. Key Concepts A radical equation is an algebraic equation in which at least one term includes a radical expression. A radical equation defines a specific relationship between numbers and variables that can be analyzed to find a solution. It is possible to find the solution to a radical equation by isolating the radical, then raising the expression to an appropriate power to eliminate the radical. Like other algebraic equations, a radical equation may have no solutions. 3.2.3: Solving Radical Equations

5. Key Concepts, continued Recall that an extraneous solution can occur when the solution process results in a value that satisfies the original equation, but creates an invalid equation; in the case of radical equations, extraneous solutions occur when a negative value results under the radical. An extraneous solution can also occur when a solution results in a value that is irrelevant in the context of the original equation. 3.2.3: Solving Radical Equations

6. Common Errors/Misconceptions raising only parts of an equation to a power when trying to eliminate a radical (for example, raising only one side of an equation to a power) confusing –x2 with (–x)2, particularly on a calculator disregarding negative signs within radical expressions neglecting to check for extraneous solutions 3.2.3: Solving Radical Equations

7. Guided Practice Example 3 Solve the radical equation for x. 3.2.3: Solving Radical Equations

8. Guided Practice: Example 3, continued Isolate the radical expression. The radical expression is already isolated on one side of the equation. 3.2.3: Solving Radical Equations

9. Guided Practice: Example 3, continued Raise both sides of the equation to a power to eliminate the radical, then simplify the result. Square both sides of the equation to eliminate the radical expression. Solve one side of the equation for 0. 3.2.3: Solving Radical Equations

10. Guided Practice: Example 3, continued Original equation Square bothsides. 2 – x = x2Simplify. 2 = x2 + x Add x to both sides. 0 = x2 + x – 2 Subtract 2 from both sides. The result is a quadratic equation. 3.2.3: Solving Radical Equations

11. Guided Practice: Example 3, continued Solve for x by factoring, or by using the quadratic formula. The quadratic appears to be factorable. 0 = x2 + x – 2 0 = (x – 1)(x + 2) The solutions appear to bex = –2 or x = 1. 3.2.3: Solving Radical Equations

12. Guided Practice: Example 3, continued Verify the solution(s). Substitute the potential solutions into the original equation, and check the result. Let’s start with x = –2. Original equation Substitute–2 for x. Simplify. 2 = –2 3.2.3: Solving Radical Equations

13. Guided Practice: Example 3, continued The result, 2 = –2, is a false statement. This solution does not satisfy the original equation. –2 is an example of an extraneous solution, one that results from the process of our solution, but does not satisfy the original equation. Check x = 1. Original equation Substitute1 for x. Simplify. 1 = 1 3.2.3: Solving Radical Equations

14. Guided Practice: Example 3, continued The result is a true statement; therefore, this solution satisfies the original equation. Since only the second solution satisfies the original equation, only that solution is valid. The solution to the radical equation is x = 1. ✔ 3.2.3: Solving Radical Equations

15. Guided Practice: Example 3, continued 3.2.3: Solving Radical Equations

16. Guided Practice Example 4 Solve the radical equation for x. 3.2.3: Solving Radical Equations

17. Guided Practice: Example 4, continued Isolate the radical expressions. The equation contains two distinct radical expressions. Each radical expression is alone on one side of the equation, so it should be possible to eliminate the radicals using the methods previously demonstrated. 3.2.3: Solving Radical Equations

18. Guided Practice: Example 4, continued Raise both sides of the equation to a power to eliminate the radicals, then simplify the result. Square both sides of the equation, then solve algebraically. 3.2.3: Solving Radical Equations

19. Guided Practice: Example 4, continued Original equation Square bothsides. x – 7 = 9(2) Simplify. x – 7 = 18 x = 25 The solution appears to bex = 25. 3.2.3: Solving Radical Equations

20. Guided Practice: Example 4, continued Verify the solution(s). Substitute the potential solution into the original equation, and check the result. Original equation Substitute25 for x. Simplify. 3.2.3: Solving Radical Equations

21. Guided Practice: Example 4, continued Since the solution satisfies the original equation, the solution is valid. The solution to the radical equation is x = 25. ✔ 3.2.3: Solving Radical Equations

22. Guided Practice: Example 4, continued 3.2.3: Solving Radical Equations