skimomma Posted January 5, 2018 Share Posted January 5, 2018 Dd is working through Intro to Geometry. Today, she was working on the exercises from section 8.6. The last exercise (8.6.6f) has a step that dd struggled with. I helped her through it but wanted to find where it was covered in the past because I am not explaining it in a way that makes sense to her. I have combed the Intro to Algebra and Pre A texts but cannot find it. In the final step, she must solve for x given the following: x^2=32-16(3)^1/2 So, she needs to factor the right side of the equation. The solution says "recognizing that 32-16(3)^1/2 = (2(6)^1/2-2(2)^1/2)^2, x=(2(6)^1/2-2(2)^1/2) Anyone know where in past chapters/texts they learned how to do this? I just *see* it, but I'm not sure how I see it and certainly am not doing a good job of explaining it to dd. Quote Link to comment Share on other sites More sharing options...
Arcadia Posted January 5, 2018 Share Posted January 5, 2018 I am thinking Prealgebra Chapter 2 exponents. 1 Quote Link to comment Share on other sites More sharing options...
JanetC Posted January 5, 2018 Share Posted January 5, 2018 My take: Messing with expressions like that is a dying art. Just put it in the calculator and be done! Quote Link to comment Share on other sites More sharing options...
Monica_in_Switzerland Posted January 5, 2018 Share Posted January 5, 2018 :bigear: I have no idea. I don't like being stumped! Quote Link to comment Share on other sites More sharing options...
Kendall Posted January 5, 2018 Share Posted January 5, 2018 It's not much of a solution to say recognizing that... I see that it is true, but I agree there needs to be an explanation of how to recognize that. I'm not real familiar with AoPS, but my daughter did PreAlg and this fall finished the exponents chapters in Intro to Alg. I don't think there was anything in there showing how to recognize that. We did not do the challenge exercises, though. Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 I am thinking Prealgebra Chapter 2 exponents. Not there! Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 It's not much of a solution to say recognizing that... I see that it is true, but I agree there needs to be an explanation of how to recognize that. I'm with you on this one! Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 I swear that I remember covering this at some point but that could be in my head. I was just sure that I was missing something obvious! (which is still possible, I realize) Quote Link to comment Share on other sites More sharing options...
JoJosMom Posted January 5, 2018 Share Posted January 5, 2018 This is of absolutely no constructive help, but here's what my daughter said: That reads like a ‘we guess and check at this point, but we’re not going to show the guess-and-check steps’. If we’ve learned how to do this, I missed it. She couldn't find it. (She's taking Intermediate Algebra.) Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 This is of absolutely no constructive help, but here's what my daughter said: That reads like a ‘we guess and check at this point, but we’re not going to show the guess-and-check steps’. If we’ve learned how to do this, I missed it. She couldn't find it. (She's taking Intermediate Algebra.) I am feeling less defeated by the post. Thanks! 1 Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 My take: Messing with expressions like that is a dying art. Just put it in the calculator and be done! I am terribly old fashion AND teach engineering students. The combo means I have never let dd use a calculator. I will. Someday. Maybe. 2 Quote Link to comment Share on other sites More sharing options...
Arcadia Posted January 5, 2018 Share Posted January 5, 2018 I swear that I remember covering this at some point but that could be in my head. I was just sure that I was missing something obvious! (which is still possible, I realize)I took a look at the AoPS Intro to geometry solution manual. The step you are referring to miss out the steps on completing the square. From my DS13 because I was too lazy to do it: (a + b (sqrt 3))^2 = 32 -16 (sqrt 3) (a^2 + 3b^2) +(2ab)(sqrt 3) = 32 -16 (sqrt 3) a^2 + 3b^2 = 32 equation (1) 2ab = -16 a = -8/b Substitute a = -8/b into equation (1) (((-8)^2)/ (b^2)) + 3b^2 = 32 8^2 -32b^2 + 3b^4 = 0 Using the quadratic formula b^2 = 8, 8/3 (skip two lines of working here) b = 2 (sqrt 2), 2 (sqrt (2/3)) a = -2 (sqrt 2), 2 (sqrt (2/3)) a + b (sqrt 3) = 2 (sqrt 2) (1 - (sqrt 3)) = 2 (sqrt 2) - 2 (sqrt 6) So 32 -16 (sqrt 3) = (2 (sqrt 2) - 2 (sqrt 6))^2 EC^2 = 32 -16 (sqrt 3) EC = 2 (sqrt 2) - 2 (sqrt 6) I need Latex enabled for this forum :P 3 Quote Link to comment Share on other sites More sharing options...
Kathy in Richmond Posted January 5, 2018 Share Posted January 5, 2018 I found a similar problem in the Intro Algebra text, Chapter 10.5. Take a look at Problem 10.30 and also Exercise 10.5.5 (same as 10.30 but no guess & check). The method of 10.5.5 is the standard approach, and that solution is just like Arcadia's solution above. (kudos to her & her son for typing all that out!) :) And take heart, 10.5.5 is a starred exercise within a starred section! 1 Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 I took a look at the AoPS Intro to geometry solution manual. The step you are referring to miss out the steps on completing the square. From my DS13 because I was too lazy to do it: (a + b (sqrt 3))^2 = 32 -16 (sqrt 3) (a^2 + 3b^2) +(2ab)(sqrt 3) = 32 -16 (sqrt 3) a^2 + 3b^2 = 32 equation (1) 2ab = -16 a = -8/b Substitute a = -8/b into equation (1) (((-8)^2)/ (b^2)) + 3b^2 = 32 8^2 -32b^2 + 3b^4 = 0 Using the quadratic formula b^2 = 8, 8/3 (skip two lines of working here) b = 2 (sqrt 2), 2 (sqrt (2/3)) a = -2 (sqrt 2), 2 (sqrt (2/3)) a + b (sqrt 3) = 2 (sqrt 2) (1 - (sqrt 3)) = 2 (sqrt 2) - 2 (sqrt 6) So 32 -16 (sqrt 3) = (2 (sqrt 2) - 2 (sqrt 6))^2 EC^2 = 32 -16 (sqrt 3) EC = 2 (sqrt 2) - 2 (sqrt 6) I need Latex enabled for this forum :p Thank You!!!! Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 I found a similar problem in the Intro Algebra text, Chapter 10.5. Take a look at Problem 10.30 and also Exercise 10.5.5 (same as 10.30 but no guess & check). The method of 10.5.5 is the standard approach, and that solution is just like Arcadia's solution above. (kudos to her & her son for typing all that out!) :) And take heart, 10.5.5 is a starred exercise within a starred section! I just KNEW I had seen this before. I cannot tell you how many times I looked at that chapter today and somehow missed this. Thanks! 1 Quote Link to comment Share on other sites More sharing options...
skimomma Posted January 5, 2018 Author Share Posted January 5, 2018 Crisis diverted! I knew I could count on the hive. 1 Quote Link to comment Share on other sites More sharing options...
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