Radical Equations and Inequalities

Transcription

1 16 LESSON Radical Equations and Inequalities Solving Radical Equations UNDERSTAND In a radical equation, there is a variable in the radicand. The radicand is the expression inside the radical symbol ( ), and the index is the root being taken. In the equation y 5 x, the radicand is x. The equation takes the square root of x, so the index is. Cube roots are also radicals (with an index of 3). UNDERSTAND A radical expression is not a polynomial, but polynomials can be used to solve radical equations. You know that linear equations are solved using inverse operations. Some quadratic equations can be solved by taking the square root of both sides. This is because raising to the nth power and taking the nth root are inverse operations. So, to solve a radical equation, raise both sides to the same power. Consider the equation x 5 3. The index is, so raise both sides to the second power. x 5 3 ( x ) 5 3 x 5 9 Since 9 5 3, the answer checks. Now consider the equation x 5 3. Solve by squaring both sides. x 5 3 ( x ) 5 (3) x 5 9 Since 9 3, the answer does not check. It is extraneous. It is important to check all answers to radical equations since extraneous solutions can occur. You should also be sure to look at the original equation and use number sense. Without solving the equation x 5 3, you could have concluded that it has no real solution, since a square root cannot be negative. 13 Unit 1: Polynomial, Rational, and Radical Relationships

2 Connect Solve: x 5 x 6 1 Square both sides of the equation. x 5 x 6 ( x ) 5 (x 6) x 5 x 1x 1 36 Solve the resulting equation for x. x 5 x 1x x 13x (x 4)(x 9) x 5 4 and x Check for extraneous solutions. Substitute each solution into the original equation. Test: x The solution x 5 4 is extraneous since it does not make the original equation true. Test: x The solution to the equation is x 5 9. TRY Solve: 3x Lesson 16: Radical Equations and Inequalities 133

3 UNDERSTAND Solving a radical inequality is much like solving a radical equation, but you will have to take some extra steps. As with other inequalities, the solution will usually be a range, which can be graphed on a number line. Consider the radical inequality x 1 3. Replace the inequality sign with an equals sign, and solve the equation. x ( x 1 3 ) 5 Square both sides. x x 5 1 Solving Radical Inequalities Subtract 3 from both sides. Now, change the equals sign back to the inequality symbol: x # 1. Graph the solution Remember that you cannot take the square root of a negative number in the set of real numbers. So, finding the domain of a radical inequality should be your first step in solving the inequality. With x 1 3, the solution only makes sense when the radicand, x 1 3, is nonnegative. This means x 1 3 $ 0. Solve this inequality for x. x 1 3 $ 0 x $ 3 Subtract 3 from both sides. Graph this solution on a number line The solution to the original inequality is the range where the two solutions overlap The solution to x 1 3 is 3 # x # Unit 1: Polynomial, Rational, and Radical Relationships

4 Connect x 1 6. x 1 Determine allowable values for the radicand. 3 DISCUSS The radicand must be greater than or equal to 0. x 1 6 $ 0 x $ 6 Check for extraneous solutions. Use the equation. Try x Try x The only solution is x 5 3. Since the original inequality has the variable on both sides, test values on either side of x to determine which inequality symbol to use (x. 3 or x, 3). The solution is x, 3. Why is 6 included in the solution to x 1 6. x, but 3 is excluded? 4 Replace the inequality sign with an equals sign, and solve the equation. ( x 1 6 ) 5 x x x 0 5 x x (x 3)(x 1 ) x 5 3 and x 5 Graph the complete solution. Graph the areas of overlap for x, 3 and x $ Lesson 16: Radical Equations and Inequalities 135

5 EXAMPLE A Solve: 3 7x x 1 Cube both sides of the equation. ( 3 7x 1 6 ) 3 5 x 3 7x x x 3 7x 6 Solve the resulting cubic polynomial. 3 Check for extraneous solutions. Test: x 5 1 ( 3 7(1) 1 6 ) Test: x 5 ( 3 7() 1 6 ) Test: x 5 3 ( 3 7(3) 1 6 ) The square root of a negative is not a real number, but the cube root of a negative number is a real number. There are no extraneous solutions. The solutions are x 5, 1, and 3. Use the rational roots theorem to find possible roots. p q 5 61, 6, 63, 66 Try 1. (1) 3 7(1) So, 1 is a root and (x 1 1) is a factor. Use synthetic division to divide by (x 1 1) Factor the resulting quadratic factor. x x 6 5 (x 1 )(x 3) Write the polynomial equation in factored form. TRY 0 5 (x 1 1)(x 1 )(x 3) The solutions are 1,, and 3. Graph the solution to 3 7x 1 6, x. 136 Unit 1: Polynomial, Rational, and Radical Relationships

6 EXAMPLE B Solve: _ x x 4 1 Determine allowable values for the radicand. The radicand must be greater than or equal to 0. x $ 0 x $ 0 The domain is x $ 0. Solve the related equation. ( _ x ) 5 (x 4) x 5 x 8x x 10x (x 8)(x ) The solutions are x 5 8 and x 5. 3 Check for extraneous solutions. Test: x 5 8? _ TRY Test: x 5? The only solution is x 5 8. Solve: 3 1 (5x 10) Test values on either side of 8 to determine the solution to the inequality. Test: x False Test: x True The complete solution is x $ 8. Lesson 16: Radical Equations and Inequalities 137

7 EXAMPLE C Solve: x x Isolate the radical expression. This will make it much simpler to square both sides. x 4 5 x 1 Solve. 3 Check for extraneous solutions. Square both sides. (x 4) 5 ( x 1 ) x 8x x 1 x 9x (x )(x 7) 5 0 The solutions are x 5 and x 5 7. Use the original equation. Test: x Test: x The only solution is x 5 7. TRY Solve: x 5 5x Unit 1: Polynomial, Rational, and Radical Relationships

8 READ Problem Solving The period of a pendulum is the amount of time required for it to swing from one side to the other and back. A pendulum s period in seconds, P, is related to the length of the pendulum in meters, L, by the following equation: P 5 p L 9.8 A pendulum in a grandfather clock has a period of.5 seconds. Find the length of the pendulum to the nearest tenth of a meter. PLAN Substitute the value of the period for P, and solve for L. SOLVE Substitute Solve for L: 5 p L 9.8 ( ) 5 ( p L 5 p for P into the equation. 9.8 ) L m CHECK Substitute the value of L into the equation that relates P and L, and solve. When L 5, P. To the nearest tenth of a meter, the pendulum is m long. Lesson 16: Radical Equations and Inequalities 139

9 Practice Identify the radicand and index of each equation. 1. y 5 3 x 1 1. y 5 4 x radicand: index: 3. 3 x radicand: index: radicand: index: 4. 1 x radicand: index: Solve. 5. 3x x REMEMBER Raise both sides of the equation to the power equal to the index x x x REMEMBER Check for extraneous solutions. 9. n n 10. n 5 11n 1 3 Choose the best answer. 11. Solve: x x 4 A. x 5 0 only B. x 5 9 only C. x 5 0 and x 5 9 D. There is no solution. 1. Solve: 3n A. x 5 1 only B. x 5 1 only C. x 5 1 and x 5 1 D. There is no solution. 140 Unit 1: Polynomial, Rational, and Radical Relationships

10 Solve. 13. x 5 x x x 15. x x x 5 x Solve. 17. A pendulum s period in seconds, P, is related to the length of the pendulum in meters, L, by the following equation: P 5 p L 9.8 A pendulum in a grandfather clock has a period of 1.7 seconds. Find its length to the nearest tenth of a meter. The pendulum s length is m. Solve each inequality. Graph the solution on the number line. 18. x x 1 6 # x 1 # x 1. 3x 1 1. x Lesson 16: Radical Equations and Inequalities 141

11 Solve.. You can estimate s, the speed in miles per hour, at which a car is moving when it goes into a skid. Use the formula s 5 1d, where d is the length of the skid marks in feet. During an accident, a driver claims to have been driving 4 miles per hour. If the driver s estimate of his speed is accurate, about how long would the skid marks be? The skid marks would be about ft long. The actual skid marks are about 115 ft. Is the driver s estimate of his speed accurate? Explain. Solve. 3. EXTEND Solve: x 1 8 x 5 Isolate the first radical. x Square both sides of the equation. x Isolate the radical expression in the resulting equation. 5 x Square both sides of the resulting equation. 5 x Show that your answer is correct by substituting it into the original equation. 14 Unit 1: Polynomial, Rational, and Radical Relationships

12 4. EXTEND Solve: 3 x x Raise both sides of the equation to the same power so that all radicals are eliminated. To do this, raise both sides to the power. Then, gather all terms on one side. 0 5 Find all real zeros of the resulting equation. The only real zero is x 5. Check the answer. Lesson 16: Radical Equations and Inequalities 143