EKS Posted September 28, 2016 Share Posted September 28, 2016 This problem came home as homework today. Algebraically solve 6 - 3x - 4x2 - 2x3 = 0 I know from graphing it on the computer that it has one real root which is irrational. I know how to find irrational roots by factoring stuff out until I get a quadratic. But I can't do that with this problem. So, how does one do this problem? (I honestly think it might be a typo. Either that or the teacher is trying to torture the students. With this particular teacher, either is possible :cursing: ) Thanks! Quote Link to comment Share on other sites More sharing options...
Janeway Posted September 28, 2016 Share Posted September 28, 2016 (edited) Google this... and you get tons of answers including the steps. Edited September 28, 2016 by Janeway Quote Link to comment Share on other sites More sharing options...
Janeway Posted September 28, 2016 Share Posted September 28, 2016 In fact, just click on my link about and you will see the Google search. Quote Link to comment Share on other sites More sharing options...
Arcadia Posted September 28, 2016 Share Posted September 28, 2016 (edited) My kid suggest using Vieta Formula with roots as a, b+ci, b-ci. Three equations and three unknowns. He is busy reading at the library or I would have ask him to work it out for you. ETA: Using Vieta's formula gave me back the original equation. I'll ask my kid to try later. ETA: Kid tried with Vieta and got back the same equation too. Edited September 29, 2016 by Arcadia Quote Link to comment Share on other sites More sharing options...
Caroline Posted September 28, 2016 Share Posted September 28, 2016 This problem came home as homework today. Algebraically solve 6 - 3x - 4x2 - 2x3 = 0 I know from graphing it on the computer that it has one real root which is irrational. I know how to find irrational roots by factoring stuff out until I get a quadratic. But I can't do that with this problem. So, how does one do this problem? (I honestly think it might be a typo. Either that or the teacher is trying to torture the students. With this particular teacher, either is possible :cursing: ) Thanks! If you change one sign, it is factorable by grouping. I vote typo. 1 Quote Link to comment Share on other sites More sharing options...
SparklyUnicorn Posted September 28, 2016 Share Posted September 28, 2016 does seem like a sign typo My instructor made 3 sign errors today rendering three problems not doable the way she wanted them done...LOL. So it happens!! Quote Link to comment Share on other sites More sharing options...
EKS Posted September 28, 2016 Author Share Posted September 28, 2016 If you change one sign, it is factorable by grouping. I vote typo. This is what I'm leaning toward as well. Quote Link to comment Share on other sites More sharing options...
EKS Posted September 28, 2016 Author Share Posted September 28, 2016 (edited) Google this... and you get tons of answers including the steps. I have... Everything I found has reducing (or whatever it's called) it to a quadratic and then using the quadratic formula. Edited September 28, 2016 by EKS Quote Link to comment Share on other sites More sharing options...
EKS Posted September 28, 2016 Author Share Posted September 28, 2016 My kid suggest using Vieta Formula with roots as a, b+ci, b-ci. Three equations and three unknowns. He is busy reading at the library or I would have ask him to work it out for you. I'll look this up. Thanks! Quote Link to comment Share on other sites More sharing options...
Guest Posted September 28, 2016 Share Posted September 28, 2016 (edited) Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms: 3(2-x)-2x^2(2-x)=0 Then factor out the (2-x) from each of the two remaining terms: (2-x)(3-2x^2)=0 Since it is equal to zero, then each of the two multiplied factors can be set to zero: 2-x=0 and 3-2x^2=0 You can probably solve it from here. OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL Edited September 29, 2016 by Kinsa Quote Link to comment Share on other sites More sharing options...
Caroline Posted September 29, 2016 Share Posted September 29, 2016 Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms: 3(2-x)-2x^2(2-x)=0 Then factor out the (2-x) from each of the two remaining terms: (2-x)(3-2x^2)=0 Since it is equal to zero, then each of the two multiplied factors can be set to zero: 2-x=0 and 3-2x^2=0 You can probably solve it from here. OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL That's why I thought it was probably a typo. Quote Link to comment Share on other sites More sharing options...
EKS Posted September 29, 2016 Author Share Posted September 29, 2016 Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms: 3(2-x)-2x^2(2-x)=0 Then factor out the (2-x) from each of the two remaining terms: (2-x)(3-2x^2)=0 Since it is equal to zero, then each of the two multiplied factors can be set to zero: 2-x=0 and 3-2x^2=0 You can probably solve it from here. OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL That's what I did at first too! Quote Link to comment Share on other sites More sharing options...
MarkT Posted September 29, 2016 Share Posted September 29, 2016 (edited) Wolfram gives x ~ 0.80121 real root x ~ -1.4006 +- 1.3352i complex roots you can display it in exact form which is a nasty looking solution - good luck - but it can be done Edited September 29, 2016 by MarkT Quote Link to comment Share on other sites More sharing options...
EKS Posted September 29, 2016 Author Share Posted September 29, 2016 Wolfram gives x ~ 0.80121 real root x ~ -1.4006 +- 1.3352i complex roots you can display it in exact form which is a nasty looking solution - good luck - but it can be done Yup. Ugh. In talking to the kid when he got home, I'm fairly sure a negative sign was left off, because they had actually done some work with factoring by grouping in class. Thanks to everyone for your help! Quote Link to comment Share on other sites More sharing options...
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