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Lesson 6.5 Summary Solving Square Roots and Other Radical Expressions Objective: To solve square root and other radical equations. MA.912.A.6.5, MA.912.A.6.4. Introduction: To solve a radical equation, isolate the radical on one side of the equation. Then raise each side to the power suggested by the index. This can introduce extraneous solutions, or solutions that do not make the original equation a true statement. A radical equation is an equation that has a variable in a radicand or a variable with a rational exponent. If the radical is 2, the equation is a square root equation. In this lesson, assume that all radicals and expressions with rational exponents represent real numbers. Notice that when you take the square root of each side of an equation, for example, you do need to introduce a ± sign. This is because the root of any even index could be either positive or negative, so in general, you introduce a ± sign when the index is even. However, when you raise both sides of an equation to either an even or odd power, you do not have to include a ± sign. 𝑚 𝑛 To solve equations of the form 𝑥 = 𝑘, raise each side of the equation 𝑛 𝑚 to the power , the reciprocal of . This is done to eliminate the 𝑚 𝑛 𝑛 𝑚 𝑚 𝑛 rational exponent. However, if either m or n is even, then (𝑥 ) = |𝑥|. This is because if you raise the radicand to an even power, it will always result in a positive number, whether the value in the radicand is positive or negative, and that positive number can have a negative and a positive root depending on whether the index is even or not. But since we are only interested in the principal, or positive, root, the value of the variable included in the expression for the root cannot be negative. On the other hand, if the index is even, the value in the radicand can have a positive and a negative root. But again, since we are only interested in the principal, or positive, root, the value of the variable included in the expression for the root cannot be negative. It is for these reasons, that the variable is represented within the absolute value sign in both cases. When solving radical equations with a rational exponent, isolate the exponential expression then use the inverse of the power to simplify and solve the equation. Isolating the radical or exponential expression is important because if you raise each side of the original equation to a power in order to remove the radical or the rational exponent you will end up with a more complicated equation, not a simpler one. Notice that when you have a radical equation whose radical expression can be converted to an exponential expression with a numerator of other than 1, it is best to convert the radical expression to an exponential expression and then raise it to the reciprocal of the rational exponent in order to remove the rational exponent. When you raise each side of an equation to a power, it is possible to introduce extraneous solutions. Therefore, it is important that you check all solutions in the original equation. An extraneous solution will give a false statement. To square a binomial, use the formula: (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 or (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 . If an equation contains two radical expressions (or two terms with rational exponents), isolate one of the radicals (or one of the terms), then eliminate it (or its rational exponent). Isolate the more complicated radical expression first. In the resulting equation, simplify the expressions before you eliminate the second radical. In lesson 6.8 you will study the graphs of square root functions. These graphs can help you find solutions and identify extraneous solutions. To graph, for example, the equation √𝑥 + 7 − 5 = 𝑥, let Y1= √ (𝑥 + 7) − 7 𝑎𝑛𝑑 𝑌2 = 𝑥. The solutions to the equation are the x values where the graphs of the two equations intercept. The extraneous solutions are those x values that could be considered as possible solutions but for which the graphs do not intercept. Demonstrated: Got It 1 pg. 391: What is the solution of √4𝑥 + 1 − 5 = 0? Classwork: ex. 13 pg. 395: Solve: √2𝑥 − 1 = 3 Demonstrated: Got It 2 pg. 392: 2 What are the solution(s) of 2(𝑥 + 3)3 = 8? Classwork: ex. 19 pg. 395: 2 3 Solve: (𝑥 + 2) = 9 Demonstrated: Got It 4a pg. 393: What is the solution of √5𝑥 − 1 + 3 = 𝑥? Classwork: ex. 27 pg. 395: 1 2 Solve: (5 − 𝑥 ) = 𝑥 + 1 Demonstrated: Got It 5 pg. 394: What is the solution of √5𝑥 + 4 − √𝑥 = 4? Classwork: ex. 38 pg. 395: Solve: √3𝑥 + 2 − √2𝑥 + 7 = 0 Homework: pgs. 395-396 ex: 14, 22, 24, 26, 37, 46, 53, 56, 64, 67.