Practice some problems before going into the exercise.
Log in Isabella 8 years agoPosted 8 years ago. Direct link to Isabella's post “is 0 an extraneous soluti...” is 0 an extraneous solution • (20 votes) Allen Yang 8 years agoPosted 8 years ago. Direct link to Allen Yang's post “It doesn't have to be. Fo...” It doesn't have to be. For example, x^2+4x=0 has solutions 0 and -4 (try them) (23 votes) Abd Hamid Mat Sain 8 years agoPosted 8 years ago. Direct link to Abd Hamid Mat Sain's post “square root 9 should be e...” square root 9 should be either 3 or -3.......why should -3 be extraneous then? • (10 votes) Kim Seidel 8 years agoPosted 8 years ago. Direct link to Kim Seidel's post “You are correct that squa...” You are correct that square root (9) can be 3 or -3. But, there is a rule that we always use the principle root (the positive value) unless the square root has a minus in front of it. See this video: https://www.khanacademy.org/math/algebra-basics/basic-alg-foundations/alg-basics-roots/v/introduction-to-square-roots (25 votes) Carla Cristina Almeida 8 years agoPosted 8 years ago. Direct link to Carla Cristina Almeida's post “In the last situation, wh...” In the last situation, why did they plug -1 in the equation? was it a random number? • (12 votes) weidner.mary62 8 years agoPosted 8 years ago. Direct link to weidner.mary62's post “This got me at first too....” This got me at first too. In the original question, it specifically asks you to use x=-1 (13 votes) Robert Roller 7 years agoPosted 7 years ago. Direct link to Robert Roller's post “So take the following as ...” So take the following as an example. sqrt(4x + 41) = x+5 Originally, x must be greater than -41/4 to be valid. When we solve this equation algebraically, we get the following two solutions: X = -8 and X = 2. When you plug X = -8 back into the equation, you end up with sqrt(9) = -3. I have three questions. • (8 votes) Himalaya 7 years agoPosted 7 years ago. Direct link to Himalaya's post “1. In your example, we us...” 1. In your example, we use the principal square root because the original problem statement says so. It says √(4x + 41), and not -√(4x + 41). By definition, the radical notation without a minus or ± sign in front of it means “the principal root”, which is always positive. The solutions to Power Equations and the solutions to Radical Equations are different things. For example, x^2 = 4 has two solutions: x = ±2, while x = √4 has only one solution: x = 2, and x = -√4 also has one solution: x = -2. We define the square root as a function, so it must have only one output for each input. If we define the function y = √x as having two solutions, then it is no longer a function. 2. The solution x = -8 is extraneous to the original equation √(4x + 41) = x + 5. However, it is the solution to the equation -√(4x + 41) = x + 5. The expression under the radical is same in both equations, so in terms of keeping the radicand non-negative, the value -8 is OK. If we take the function y = √(4x + 41), then -8 would be a valid input value for x. However, for the equation √(4x + 41) = x + 5, the value x = -8 is not a solution, because it leads to an invalid statement 3 = -3, which is not true. 3. As mentioned above, x = -8 is the solution to the equation -√(4x + 41) = x + 5. So, if we graph -√(4x + 41) instead of √(4x + 41), it will intersect with x + 5 at x = -8. (12 votes) 1earth4ever 6 years agoPosted 6 years ago. Direct link to 1earth4ever's post “I am having a really hard...” I am having a really hard time with this unit. I have watched all the videos several times and I am still very confused. I would really appreciate some help. • (7 votes) Ian Pulizzotto 6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “The key idea for solving ...” The key idea for solving square root equations is to isolate a radical on one side and square both sides. If there's still a radical in the equation, then this process would need to be performed a second time. (By the way, don't forget to include the middle term when squaring a binomial. Many students forget to do this.) After you have solved the resulting linear or quadratic equation for x, remember that you're not finished yet! Because every positive number has a positive and a negative square root, but the radical symbol denotes only the positive (principal) square root, the act of squaring both sides can create invalid (extraneous) solutions! So plug in your solutions to the original equation to determine which solutions work and which solutions must be discarded. Have a blessed, wonderful day! (10 votes) Tetsuya 7 years agoPosted 7 years ago. Direct link to Tetsuya's post “I still don't understand ...” I still don't understand why I should care about extraneous solution. It's outside the domain, not a solution, a wrong answer. Or is there any use of finding it? • (6 votes) kubleeka 7 years agoPosted 7 years ago. Direct link to kubleeka's post “Extraneous solutions are ...” Extraneous solutions are not necessarily outside the domain. But they can appear as extra solutions when we square both sides of an equation, because when we square an equation, we would get the same result whether the original equation was positive or negative. So one solution corresponds to the positive equation, and the other to the negative equation. But both of them will fall out of the algebra. We need to be able to tell which solution is extraneous and which works. (10 votes) Sally 4 years agoPosted 4 years ago. Direct link to Sally's post “Hi I don't understand why...” Hi I don't understand why √9 would not equal -3 when it equals 3. In practice question 3, why wouldn't x=-1 be a correct solution?? • (5 votes) Esther Hernandez 4 years agoPosted 4 years ago. Direct link to Esther Hernandez's post “I think they're looking f...” I think they're looking for the principal sqrt of 9. (5 votes) Joslynkim75 5 years agoPosted 5 years ago. Direct link to Joslynkim75's post “In the video titled, "Equ...” In the video titled, "Equation that has a specific extraneous solution." The person who did the video squared the right side of the equation wrong. (d+2x)^2 should be d^2 + 4dx + 4x^2. This is an error. • (4 votes) Kim Seidel 5 years agoPosted 5 years ago. Direct link to Kim Seidel's post “This is a known error. A...” This is a known error. A correction box does pop up on the video at about (5 votes) Courtney Ellwein a year agoPosted a year ago. Direct link to Courtney Ellwein's post “In question 4 why can't t...” In question 4 why can't the square root of 9 = -3 ? • (3 votes) Kim Seidel a year agoPosted a year ago. Direct link to Kim Seidel's post “The problem tells you tha...” The problem tells you that it wants the principal root (the positive root). If the problem wanted the negative root, it would have a "-" in front. See: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rational-exponents-radicals/x2f8bb11595b61c86:radicals/v/introduction-to-square-roots (7 votes) Shyawn Zahid 8 years agoPosted 8 years ago. Direct link to Shyawn Zahid's post “For the question with Ele...” For the question with Elena, doesn't she have to check too because it's possible to have no solutions? • (3 votes) Denis Son 8 years agoPosted 8 years ago. Direct link to Denis Son's post “we squared the both sides...” we squared the both sides by an odd number. So it's not necessary to check the equation (2 votes)Want to join the conversation?
sqrt(9) is asking for the principle root = 3
- sqrt(9) is asking for the negative root = -3.
This is extraneous because we are supposed to use the principle root.
#1: Why do we have to use the principle root? It seems logical that the sqrt(9) should be equal to both +3 and -3.
#2: Why is this extraneous solution still within the original domain of the function, as it is > -41/4.
#3: What happened mathematically to produce this extraneous solution. If you graph the two equations separately, they only intersect at x = 2, therefore x = 8 is not a solution. But why is this? What happened to produce this extra answer?
1:05
sqrt(9) = +3, the principal root
-sqrt(9) = -3, the negative root
