xLeeChaex 46 Posted October 14, 2014 Share Posted October 14, 2014 I wanna know for number 16, my teacher told me the answer for it is no solution when using the elimination method for it I solved it but not sure how its no solution though..i think the way i solved it is right though lol can u please help? its a review for my test 2morrow XD Link to comment Share on other sites More sharing options...
Jon Snow 14,255 Posted October 14, 2014 Share Posted October 14, 2014 oh taking the square root of negative numbers/variables result in imaginary numbers aka no solution imaginary numbers like 5i Link to comment Share on other sites More sharing options...
jvng 3,864 Posted October 14, 2014 Share Posted October 14, 2014 and here i am struggling with that pythagoras bullshit Link to comment Share on other sites More sharing options...
27lines 781 Posted October 14, 2014 Share Posted October 14, 2014 I think there's a sign error. Right after you find x=2 as a candidate you use x=-2. x=2 and y=(+/-)sqrt(2) are definitely solutions (check!) and, indeed, the only solutions in the real numbers. Link to comment Share on other sites More sharing options...
Chipo 60 Posted October 14, 2014 Share Posted October 14, 2014 x = 2 works fine; you've just mixed up the signs: y2 = 2 not -2 Link to comment Share on other sites More sharing options...
xLeeChaex 46 Posted October 14, 2014 Author Share Posted October 14, 2014 I think there's a sign error. Right after you find x=2 as a candidate you use x=-2. x=2 and y=(+/-)sqrt(2) are definitely solutions (check!) and, indeed, the only solutions in the real numbers. I see but it will still be square root -2 meaning no solution I believe... Link to comment Share on other sites More sharing options...
abra 40,910 Posted October 14, 2014 Share Posted October 14, 2014 What does that crossed over O or 0 mean? Link to comment Share on other sites More sharing options...
Chipo 60 Posted October 14, 2014 Share Posted October 14, 2014 Those empty signs, inverted Øs or whatever are supposed to be zero, right? so if x - y2 = 0 and x = 2 then you have 2 - y2 = 0 --> 2 = y2 --> y = sq root of 2 Link to comment Share on other sites More sharing options...
♔Nippon Girl♥ 737 Posted October 14, 2014 Share Posted October 14, 2014 even thought I am good at math I don't wanna use my brain sorry Link to comment Share on other sites More sharing options...
Athena 21,718 Posted October 14, 2014 Share Posted October 14, 2014 I see but it will still be square root -2 meaning no solution I believe... No it will be just root 2, not root -2. 2 - y^2 = 0 -2 -2 __________ -y2 = -2 -----> divided by -1 -------> y2 = 2 -------> y = root 2 Those should be considered as solutions though right? Unless your teacher only wants integers for the solution? ------- For the other variable, -3, your teacher is correct because square root of -3 = root 3 times i which is imaginary. Link to comment Share on other sites More sharing options...
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