cintinative Posted October 26, 2021 Share Posted October 26, 2021 (edited) I have the answer but not a mapped out solution. I am coming up with -5/4 and -5 but they only have -5/4 listed as an answer in the key. ETA: I forgot to test both answers in the original equation. Rookie mistake. sorry guys. Edited October 26, 2021 by cintinative Quote Link to comment Share on other sites More sharing options...
purpleowl Posted October 26, 2021 Share Posted October 26, 2021 (edited) If you use -5, then you would have to take the negative roots in order for it to work in the initial equation. With positive roots (which is what is assumed when no sign is given), you get 4=6, which doesn't work. So -5 is an extraneous solution. Edited October 26, 2021 by purpleowl 1 Quote Link to comment Share on other sites More sharing options...
cintinative Posted October 26, 2021 Author Share Posted October 26, 2021 (edited) thank you @purpleowl! I completely forgot to check the answers in the original equation. doh. Edited October 26, 2021 by cintinative 2 Quote Link to comment Share on other sites More sharing options...
EKS Posted October 26, 2021 Share Posted October 26, 2021 Does anyone else think that there should be a law against using s as a variable? 2 1 Quote Link to comment Share on other sites More sharing options...
purpleowl Posted October 26, 2021 Share Posted October 26, 2021 Just now, EKS said: Does anyone else think that there should be a law against using s as a variable? YES! 1 Quote Link to comment Share on other sites More sharing options...
cintinative Posted October 26, 2021 Author Share Posted October 26, 2021 50 minutes ago, EKS said: Does anyone else think that there should be a law against using s as a variable? Yes! The first thing I did was replace it with x. My s's look like my 5s Quote Link to comment Share on other sites More sharing options...
daijobu Posted October 26, 2021 Share Posted October 26, 2021 When you square both sides of an equation, you risk introducing extraneous solutions, so you need to test all your solutions in the original equation. Plugging in into the original equation we get the left hand side, and the right hand side, , so this is extraneous. Trying we get both sides of the original equation, , so this is our valid solution. 1 Quote Link to comment Share on other sites More sharing options...
cintinative Posted October 26, 2021 Author Share Posted October 26, 2021 2 minutes ago, daijobu said: When you square both sides of an equation, you risk introducing extraneous solutions, so you need to test all your solutions in the original equation. Plugging in into the original equation we get the left hand side, and the right hand side, , so this is extraneous. Trying we get both sides of the original equation, , so this is our valid solution. Yes, I just totally forgot to do this. Thanks for the reminder! Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.