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Two-value system with square roots

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a problem with lines in analytic geometry, and I solved it in a certain way (parallel lines interceptions) which gives the correct result, and I'm happy with that.

    There was another method I thought I could use to solve it though, which went through the formulas of distance between two points. Now, I set up the equations, and I get a system with two variables (x and y) and two equations. I tried verifying the equations by replacing the variables with the results I already know, and the equations seem to be correct. The problem is, I don't know how to actually solve the system because of a square root on the way.

    So here is the system I get:

    1. (x+3)2 + (3-y)2 = 25
    2. (1-x)2 + y2 = 20

    The results should be x = -3; y = -2

    3. The attempt at a solution
    I tried getting y2 from the second equation, but I also need y (as in y1) if I want to replace it in the first equation, due to that -6y. So the first equation becomes:

    x+9+6x+9-25+20-1-x2+2x -6*√(19+2x-x2)

    I also tried subtracting the second equation from the first equation, expecting not to work, and indeed this is as far as I can get:

    8x -6y + 12 = 0

    So, how I am supposed to solve this system?
     
  2. jcsd
  3. Apr 18, 2017 #2

    PeroK

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    That last equation looks good. That gives you ##x## in terms of ##y##. Why not put that back into one of your original equations?
     
  4. Apr 18, 2017 #3

    Mark44

    Staff: Mentor

    This is not an equation. Is there a typo here?
     
  5. Apr 19, 2017 #4
    That was = 0, my bad :p

    I thought that would just give me back the original equations again. Checked it, and it doesn't lol In fact, it gives the right results, so thanks.

    It actually gives two results: ##y## = -2 (the right one), and ##y## = 22/5 (which is geometrically unacceptable for what the problem was originally asking).

    Although I solved it now, there's something I'd like to ask on this same topic (squared systems). I actually managed to get the right results in another way (in my method, I also got -2 and 22/5 for ##y##), but I'd like to know if it is technically correct (albeit much harder than it had to be).

    1. The first thing I did was squaring the first equation, thus getting:
    ##y##2 - 6##y## + (##x##2 + 6##x## -7) = 0

    2. Solved the previous equation for ##y## -> (various steps) -> ##y## = 3 ± ##\sqrt{- (x+8)(x-2)}##

    3. Plugged the result I found from the first equation into the second equation, thus getting:
    -8##x## +6 ± 6*##\sqrt{- (x+8)(x-2)}## = 0

    4. Then I moved ± 6*##\sqrt{- (x+8)(x-2)}## to the right, and from now on I consider two separate equations individually:
    -8##x## +6 = - 6*##\sqrt{- (x+8)(x-2)}##
    And:
    -8##x## +6 = + 6*##\sqrt{- (x+8)(x-2)}##

    5. Now here's the step I'm unsure about: I squared everything (also, I just realized I could have divided by two before doing this. Anyway).

    (-8##x## +6)2 = (- 6*##\sqrt{- (x+8)(x-2)})## 2
    (-8##x## +6)2 = (+ 6*##\sqrt{- (x+8)(x-2)})## 2

    I know that you can't just square stuff (unless you know they have the same sign left and right, I think; or maybe it's when they are both positive? Anyway, definitely not when you don't know the signs), but I tried with a simpler example to see if this could somehow work.

    6. This is the example I used:
    (##x##-1) = (-3)
    (##x##-1)2 " = " (-3)2 -> (...) -> ##x## = +4; ##x## = -2

    So I deduced that when you square without knowing the sign, you will get a new equation which gives one true result, and one "false" result (##x## = -2 would be the true result, which gives -3 (-2-1 = -3); ##x## = +4 would give +3 instead (+4-1 = +3)). I applied this deduction to my problem, assuming it's a general thing that always works.

    So I tried it, replaced the "##x##" results I got (after squaring, so that's two for each equation) from step 5 to verify the equations back in step 4, and indeed I got something similar: one of the results would backward-verify the equation, while the other result wouldn't (just like ##x## = -2 verified the equation, but ##x## = +4 didn't).

    So my question is: in a general case, where you have something like:
    (##x##-1) = -3

    If you wanted to, is it mathematically correct (I know it's formally ugly and unnecessary, but I care about whether or not this would always work, at least with Real numbers) to square everything, and then "backwards-check" which of the two solutions you get actually satisfies the original equation (before squaring), and which one doesn't? Or did I just get lucky, picking an easy example where this works, and then lucky again because with the equation I had in my problem this worked again?

    It's not the first time I mess around with numbers and end up with an equation where ##x## is in a square root. I usually find a way around it, but I'd really like it if the method I used here is actually correct (albeit, again, unnecessary). In other words, the next time I see something like this:

    x + ##\sqrt{4x^2+9}## = 7
    Or this:
    x + ##\sqrt{8x+9}## = 7

    Can I, while not violating any mathematical principle, proceed by squaring everything in a convenient way, and then backward-check the results?
    I tried these last two equations with my method; the first one yielded two acceptable solutions (wolfram alpha confirmed), while the second one yielded x = 2 (backwards-acceptable) and x = 20 (not backwards-acceptable).
     
  6. Apr 19, 2017 #5

    PeroK

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    Yes, that's what you have to do if you simplify an equation by squaring it. The simplest example is:

    ##x = 1 \ \Rightarrow \ x^2 = 1 \Rightarrow \ x = \pm 1##

    In mathematical logic, this is called a one-way implication, where you can go logically from left to right, but not necessarily back from right to left. It means you may have solutions to the equation on the right that aren't solutions to the original equation. You then need to check each solution.

    You could compare this with other simplifications where the implication goes both ways:

    ##x + 1 = 2 \Leftrightarrow x = 1##

    In this case, as the logic goes both ways, you know that the equation on the right is equivalent to the one on the left; and, any solution to the equation on the right must also be a solution to the equation on the left.
     
  7. Apr 20, 2017 #6
    Perfect. That'll probably come in handy in the future. Thanks a lot.
     
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