1. 1 PROBLEM 1 Standard 15 PROBLEM 2 SOLUTION OF RADICAL EQUATIONS AND INEQUALITIES PROBLEM 3 PROBLEM 4 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 STANDAR 15: Students determine whether a specific algebraic statement involving rational expressions, radical expressions, or logarithmic or exponential functions is some-times true, always true, or never true. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 ESTÁNDAR 15: Los estudiantes determinan si un estatuto algebráico involucrando expresiones racionales, expresiones radicales o funciones logarítmicas o exponenciales es algunas veces cierto, siempre cierto, o nunca cierto. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 Standard 15 Solve: y -6 =y 2 +6 y = y 2 + 6 y - y 2 = 6 y(1 - 2 ) = 6 (1 - 2 ) ( 1 - 2 ) y = 6 (1 - 2 ) 1 + 2 y = 6( ) 1 - 2 1 + 2 2 2 y = 6 + 6 2 1 – 2 y = 6 + 6 2 y = -6 -6 2 Check: y -6 = y 2 -6 – 6 2 -6 = (-6 – 6 2 ) 2 -12 – 6 2 = -6 2 - 6 ( 2 ) 2 -12 – 6 2 = -6 2 - 12 The solution is -6 – 6 2 -y 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 Standard 15 Solve: k – 6 = 3 - k 2 2 k – 6 = 3 - 2 3 k + k 2 2 k – 6 = 3 – 2 3k + k -k -3 -9 = - 2 3k 2 2 81 = 4(3k) 81 = 12k 12 k= 81 12 81 27 9 3 1 3 3 3 3 12 6 3 1 2 2 3 GCF = 3.... 3 3 27 4 = Check: k – 6 = 3 - k – 6 = 3 - 27 4 4 – 6 = 3 - 27 4 4 4 3 2 2 2 – = 3 - 27 4 24 4 3 2 3 4 = 3 (1 - ) 3 2 3 - 3 2 = The solution doesn’t check; so, there is no real solution! 81 =-2 3k 2 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. Standard 15 Solve: 3(2 –17x) + 6 = 0 1 5 -6 3(2 –17x) = -6 1 5 3 3 (2 – 17x) = -2 1 5 5 5 (2 – 17x) = -32 5 5 2 – 17x = -32 -2 -17x = -34 -17 x = 2 3(2 –17x) + 6 = 0 1 5 Check: 3(2 –17(2)) + 6 = 0 1 5 3(2 –34) + 6 = 0 1 5 3(-32) + 6 = 0 1 5 3(-2) + 6 = 0 -6 + 6 = 0 0 = 0 The solution is x = 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. Standard 15 Solve: 4x +4 + 6 12 < First find out admissible values for the radical expression: 4x + 4 0 > -4 4x -4 > 4 x -1 > Now solve the inequality: 4x +4 + 6 12 < -6 4x +4 6 < 2 2 4x +4 36 < -4 4x 32 < 4 4 x 8 < x 4 2 6-2 -4-6 8 10 -8 -10 0 -1 x 8 < < Checking the solution with a value in the 3 areas of the graph: 4(-2) +4 + 6 12 < 4(0) +4 + 6 12 < 4(9) +4 + 6 12 < -4 + 6 12 < 4 + 6 12 < 8 12 < 40 + 6 12 < 12.3 12 < There is no real square root for a negative number: false false The area of the solution has values that are true. true PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 Standard 15 To solve inequalities with radicals: 1.Check the restricted values for the expressions within radicals: remember there is no real value for a negative value of an even root. 2.Solve the inequality. 3.Check the solution set with values in all areas in and out the solution set. The solution set is where those values yield a true result. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved