   sqrt(1)!=-1?
  Okay, I have a quiz Wednesday and I still don't understand this...Lookie here..sqrt(2x-1) = x - 2solve get:x = 5 || x = 1using quadratic: under the sqrt() you get b^2-4aca=1, b=-6, c=5or sqrt(16) or 4doing the rest ((6) (+-) 4)/2(1)+:(6 + 4)/2 = 5-:(6 - 4)/2 = 1My professor says that sqrt(2(1)-1) != (1)-2 orsqrt(1) != -1Why not?  He mentions that squaring both sides to solve creates a false answer, but as far as I know, (-1)(-1) = 1.  Please help me understand.   
  (-1)(-1) does equal 1, however when you square both sides of an equation it may introduce extraneous solutions.  By your same logic, -1 == 1 because -1^2 == 1^2.Edit: So basically, always check your solutions in the original equation.   ;  Wow, you sound just like my professor but you explained a bit better...So...just another question..if you do the same to both sides, it still remains truex==xx+2==x+2orx == 6x + 2 == 6 + 2why is it different with squares (^2) than with that simple addition (+)?I would like to say, I am sorry if I am sounding dense, but maybe another example could help?Edit: crap, it just hit me like a ton of bricks...thanks for the response.   ;Quote:Original post by ImgellingI would like to say, I am sorry if I am sounding dense, but maybe another example could help?Here's a proof that 3 = 4:3 = 43*0 = 4*00 = 0So obviously 3 = 4!   ;Quote:Original post by juturnas(-1)(-1) does equal 1, however when you square both sides of an equation it may introduce extraneous solutions.  By your same logic, -1 == 1 because -1^2 == 1^2.Edit: So basically, always check your solutions in the original equation.Okay, so, I am still confused the more I think about your answer.  You start with a false equation.By your logic based on mine (and a partridge in a pear tree) :3 = 5 // false3^2 = 5^2  // false, no crapbut what if I gave you as a math test question:sqrt(1) = -1you would find that to be true right?I am not trying to be dense, I am really not, and this might be an obscure rule that I don't remember.But really, if I presented you with:sqrt(1) = -1you would equate that to be true right? Or am I missing the whole point.  Once again, I am sorry for being annoying.;Quote:Original post by SneftelQuote:Original post by ImgellingI would like to say, I am sorry if I am sounding dense, but maybe another example could help?Here's a proof that 3 = 4:3 = 43*0 = 4*00 = 0So obviously 3 = 4!Once again (you replied as I was), you started with a false statement.edit: if I am missing something obvious, let me know please!   ;Quote:Original post by ImgellingOnce again (you replied as I was), you started with a false statement.It's the same thing, however. By starting with sqrt(1) = -1, you're also starting with an incorrect assumption, and then using the "power" operator to "prove" your incorrect assumption is correct.So, you can say:sqrt(1) = -1(sqrt(1))2 = (-1)21 = 1Or you can say:3 = 43*0 = 4*00 = 0They both start with an incorrect assumption and then proceed to "prove" that it's "true".   ;Quote:but what if I gave you as a math test question:sqrt(1) = -1you would find that to be true right?No. sqrt is a function. A function, by definition, has only one possible output for a given input. sqrt(1) cannot be -1, because it's already busy being 1.Of course, -1 IS a square root of 1. Nevertheless, the sqrt function returns the positive square root of the number.   ;Quote:but what if I gave you as a math test question:sqrt(1) = -1you would find that to be true right?No. sqrt is a function. A function, by definition, has only one possible output for a given input. sqrt(1) cannot be -1, because it's already busy being 1.Of course, -1 IS a square root of 1. Nevertheless, the sqrt function returns the positive square root of the number.   ;Quote:Original post by SneftelQuote:but what if I gave you as a math test question:sqrt(1) = -1you would find that to be true right?No. sqrt is a function. A function, by definition, has only one possible output for a given input. sqrt(1) cannot be -1, because it's already busy being 1.Of course, -1 IS a square root of 1. Nevertheless, the sqrt function returns the positive square root of the number.Sneftel:I'm sorry, I see your point with the function, but I didn't know how to draw (write) the sqrt() text.  I didn't mean it as a function with programming.  Sorry for the misunderstanding.Codeka:1.  How are you writing super/sub script?2.  What is the incorrect assumption in the fact that the square root of 1 = {1,-1}?  I assume the fact of -1 * -1 = 1 and 1 * 1 = 1, both true right?  Therefore (can't make that symbol either), the square root of 1 = {1,-1}. Basically, should I just realize, as a rule, that if the starting equation has a variable on both sides and when I solve and it becomes a square root of a number that equals a negative number, I should answer that number is not a solution?Once again, sorry, I am not trying to be difficult, just trying to understand.Edit: I think my main problem is the fact the numeral 1 comes up as a possible answer and just throwing my world in a spin.   