Teacher Road Map for Lesson 23: Solving Complicated Radical Equations

Transcription

1 eacher Road Map for Solving Complicated Radical Equations Objective in Student Friendly Language: oday I am: comparing two methods for solving complicated radical equations So that I can: determine the steps to solve radical equations. I ll know I have it when I can: apply this skill to a geometry problem. Flow: o Students compare two methods for solving complicated radical equations. o Students solve four equations and use them to write the steps for solving radical equations. o Students look at situations where the radical cannot be isolated. o Students apply their skills to a geometry problem. Big Ideas: o Radical equations become polynomial equations through exponentiation. CCS Standards: o A-REI.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, startingfrom the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. o A-REI. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. 33

2 Student Outcomes Students develop facility in solving radical equations. Lesson Notes In the previous lesson, students were introduced to the notion of solving radical equations and checking for extraneous solutions (A-REI.). Students continue this work by looking at radical equations that contain variables on both sides. he main point to stress to students is that radical equations become polynomial equations through exponentiation. So we really have not left the notion of polynomials that have been studied throughout this module. his lesson also provides opportunities to emphasize MP.7 (look for and make use of structure). In the last lesson you solved equations with radicals but there was only one variable term. In this lesson, you ll expand your equation-solving skills to include equations with radical and non-radical variable terms. Opening Exercise (15 minutes) 1. Carlos and Andrea were solving the equation x + x= 0. Andrea says that there are two solutions, 0 and -. Carlos says the only solution is - because he divided both sides by x and got x + = 0. Who is correct and why? Andrea is correct. If we replace x with either 0 or, the answer is 0. When Carlos divided both sides of the equation by x, he changed the solutions from 0 and to simply. He lost one solution to the equation. Performing algebraic steps may alter the set of solutions to the original equation.. Carlos and Andrea are solving a different equation x = 3. Andrea says the solution is 9 because she squared both sides and got x = 9. Carlos says there is no solution. Who is correct? Why? Andrea is correct when she says to eliminate a radical from an equation, we raise both sides to an exponent, but in this case her answer is extraneous. If we let x = 9, then we get 9 = 3, and 3 3, so 9 is not a solution. he danger in squaring both sides of an equation is that it sometimes produces an equation whose solution set is not equivalent to that of the original equation. If both sides of x = 3 are Scaffolding Use several examples to illustrate that if aa > 0, then an equation of the form xx = aa will not have a solution (e.g., xx = 4, xx = 5). Extension: Write an equation that has an extraneous solution of xx =

3 squared, the equation x = 9 is produced, but 9 is not a solution to the original equation. he original equation has no solution. Because of this danger, the final essential step of solving a radical equation is to check the solution or solutions to ensure that an extraneous solution was not produced by the step of squaring both sides. We could have predicted that the equation would have no solution since the square root of a number is never equal to a negative value, so there is no x-value so that x = 3. While this problem is difficult, students should attempt to solve it on their own first, by applying their understandings of radicals. Students should be asked to verify the solution they come up with and describe their solution method. Discuss Exercise 3 as a class once they have worked on it individually. 3. Carlos and Andrea tried one more problem, 6 = x+ x. hey both agreed that you need to square both sides, but Andrea wants to isolate the x first and Carlos thinks they should isolate the x. Finish both methods and then decide which you like best. Do they both give the same answer? ( 6 x) Carlos Method 6 = x+ x 6 x = x 36 1 x + x= x = x = ( 1 x) = ( x x 36) 1 x x x x= x x 71x + 7x = x x 71x 7x+ 196 (Note: at this point Carlos needs to know a couple of the factors of this polynomial to continue. 9 and 4 are both factors and can be used to rewrite this equation in factored form.) Andrea s Method 6 = x+ 6 x= x ( 6 x) = ( x) 36 1x+ x = x x 13x+ 36 = 0 ( x 9)( x 4) = 0 x= 9; x= 4 Check: 6-9=-3 9 = 3 Check: 6-4= = 4 here is one extraneous solution (x = 9) and one valid solution (x = 4). x 35

4 Discussion Questions How does this equation differ from the ones from yesterday s lesson? here are two xx s; one inside and one outside of the radical. Explain how you were able to determine the solution to the equation above. Isolate the radical and square both sides. Solve the resulting equation. Did that change the way in which the equation was solved? Not really; we still eliminated the radical by squaring both sides. What type of equation were we left with after squaring both sides? A quadratic polynomial equation Why did 9 fail to work as a solution? Exercise 4 (13 minutes) he square root of 9 takes only the positive value of 3. Allow students time to work the problems independently and then pair up to compare solutions. Use this time to informally assess student understanding by examining their work. Display student responses, making sure that students checked for extraneous solutions. he steps students write may vary. Discuss as a class. 4. Solve each radical equation and then write the steps needed to solve radical equations. A. 33xx = 11 + xx he only solution is 1. Note that 1 is an extraneous solution. 9 B. 33 = 44 xx xx he two solutions are 9 and 1. C. xx + 55 = xx 11 he only solution is 44. Note that 11 is an extraneous solution. D xx 88 = 00 Steps to Solve Radical Equations 1. If there are two radicals, then have one on each side of the equation. If there is only one radical, then isolate it as much as possible.. Square both sides of the equation. 3. If there were two radicals in the original equation, you ll still have one radical after squaring. Isolate that radical and square both sides again. 4. You should now have a polynomial equation. Get all the terms on one side of the equation so that your equation is equal to Factor then solve. 6. Check for extraneous solutions. here are no solutions. 36

5 Exercise 5 (5 minutes) What if there is no way to isolate the radical? What is going to be the easiest way to square both sides? 5. Solve the equation xx + xx + 33 = 33. MP.7 Give students time to work on Exercise 5 independently. Point out that even though we had to square both sides twice, we were still able to rewrite the equation as a polynomial. Exercise 5 Solve the equation xx + xx + 33 = 33. Scaffolding: What if we had squared both sides of the equation as it was presented? his is similar to Carlos method in the Opening Exercise. xx + 33 = 33 xx xx + 33 = (33 xx) xx + 33 = xx + xx 11 = xx 11 = xx Check: = 11 + = 33 So the solution is 11. Exercise 6 (7 minutes) Allow students time to work the problems independently and then pair up to compare solutions. Circulate to assess understanding. Consider targeted instruction with a small group of students while others work independently. Display student responses, making sure that students check for extraneous solutions. 6. Solve the following equations. A. xx 3 + xx + 5 = 4 x = 4 B. 3 + xx = xx + 81 x =

6 7. Consider the triangle AAAAAA shown to the right where AAAA = DDDD, and BBBB is the altitude of the triangle. A. If the length of BBBB is xx cccc, and the length of AAAA is 1111 cccc, how long is AD and DC? 9 cm each B. Write an expression for the lengths of AAAA and BBBB in terms of xx. (Hint: Use the Pythagorean heorem.) AAAA = BBBB = 81 + xx cm C. Write an expression for the perimeter of AAAAAA in terms of xx xx + 18 cm D. Find the value of xx for which the perimeter of AAAAAA is equal to 3333 cccc. 19 cm Lesson Summary If a = b and n is an integer, then a n = b n. However, the converse is not necessarily true. he statement a n = b n does not imply that a = b. herefore, it is necessary to check for extraneous solutions when both sides of an equation are raised to an exponent. Example: (-3) = 3 but

7 Closing (5 minutes) Ask students to respond to these prompts in writing or with a partner. Use these responses to informally assess their understanding of the lesson. How did these equations differ from the equations seen in the previous lesson? Most of them contained variables on both sides of the equation or a variable outside of the radical. How were they similar to the equations from the previous lesson? hey were solved using the same process of squaring both sides. Even though they were more complicated, the equations could still be rewritten as a polynomial equation and solved using the same process seen throughout this module. Give an example where aa nn = bb nn but aa bb. We know that ( 3) = 3 but 3 3. Exit icket (5 minutes) 39

8 Name Date Solving Radical Equations Exit icket 1. Solve xx + 15 = xx + 6. Verify the solution(s).. Explain why it is necessary to check the solutions to a radical equation. 330

9 Exit icket Sample Solutions 1. Solve xx = xx Verify the solution(s). he solutions are 33 and 77. Check xx = 33: ( 33) = 99 = = 33 So, 33 is a valid solution = xx = xx = (xx + 33)(xx + 77) Check xx = 77: herefore, the only solution to the original equation is 33. ( 77) = 11 = = 11 Since 11 11, we see that 11 is an extraneous solution.. Explain why it is necessary to check the solutions to a radical equation. Raising both sides of an equation to a power can produce an equation whose solution set is not equivalent to that of the original equation. In the problem above, xx = 77 does not satisfy the equation. Homework Problem Set Sample Solutions 1. Solve. A. xx 55 xx + 66 = B. xx 55 + xx + 66 = 00 No solution C. xx 55 xx + 66 = No solution D. xx 55 xx + 66 = 4444 E. xx + 44 = 33 xx 3333 F. xx + 44 = 33 + xx No solution 331

10 G. xx + 33 = 55xx H. xx + 11 = xx I. xx xx = K. xx = 44xx J. xx = 11 44xx L. 44xx 11 = xx 11 M. xx + = 44 xx 66 N. xx xx 1111 = O. xx = xx , 88 P. xx = 99xx , 88. Consider the right triangle AAAAAA shown to the right, with AAAA = 88 and BBBB = xx. A. Write an expression for the length of the hypotenuse in terms of xx. AAAA = 64 + xx B. Find the value of xx for which AAAA AAAA = 99. he solutions to the mathematical equation 64 + x 8 = 9 are 15 and 15. Since lengths must be positive, 15 is an extraneous solution, and x =