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A strange solution ?

  1. Oct 2, 2009 #1
    a strange solution ??????

    I was solving a tangent problem and came accross a strange thing ,
    its a simple quadratic equation ,y2 - 4y -8 = 0 ,
    by quadratic formula , y = 2 + 2*31/2 and 2 - 2*31/2 ,

    now if i write the same quadratic as (y-2)2 = 12 ,
    then i apply square root function on both sides to get y-2 = (12)1/2
    NOTE : i have not written +,- in front of root 12 because on left side i have a +ve number as it is a square so then i get the answer y = 2*(3)1/2 + 2
    now how can we explain this?????????
     
  2. jcsd
  3. Oct 2, 2009 #2
    Re: a strange solution ??????

    Your mistake is in taking the square root of a square to "undo" it. When x is negative, it is a blatant lie to say [tex]\sqrt{x^2} = x[/tex]. When x is negative, [tex]\sqrt{x^2} = -x[/tex]. Try it out for yourself.

    How can you avoid this mistake. Answer: always keep in mind when you root a square. Undoing a square with a root is technically an illegal move unless you know whether x is positive.

    How do you legalize this kind of move? You break it into two cases. Draw a vertical line down the middle of your paper. On the left hand side, make the assumption that x is negative. Since every real number must be negative or non-negative, one of the two halves of the paper must logically follow. At the bottom, we get something like y = .... on either side of the page, and those are our solutions.

    To illustrate:

    [tex]y^2 - 4y - 8 = 0[/tex]
    [tex]y^2 - 4y + 4 = 12[/tex]
    [tex](y - 2) ^2 = 12[/tex]

    Case 1: Suppose [tex]y - 2 \ge 0[/tex].

    [tex]y - 2= \sqrt{12}[/tex]
    [tex]y = \sqrt{12} + 2[/tex]

    Case 2: Suppose [tex]y - 2 < 0[/tex].

    [tex]y - 2= -\sqrt{12}[/tex] (note the minus sign on the RHS)
    [tex]y = 2 -\sqrt{12}[/tex]

    So our two possible solutions are [tex]y = \sqrt{12} + 2[/tex] or [tex]y = 2 - \sqrt{12}[/tex].

    This is probably the most common mistake every single student of algebra makes.
     
    Last edited: Oct 2, 2009
  4. Oct 2, 2009 #3

    HallsofIvy

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    Re: a strange solution ??????

    Or: [itex]\sqrt{x^2}= |x|[/itex], not just "x".
     
  5. Oct 2, 2009 #4

    jambaugh

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    Re: a strange solution ??????

    It helps to remember that "a solution" does not mean the result of your solving process but a value that makes the original equation true.

    This is why we teachers keep harping on "check your answers!". Some mistakes are of the 2+2=5 variety but others are more subtle like this case where you are using a method outside the simpler context in which you originally learned it.

    This issue comes up again and again e.g. inverse sine of sine of 7pi/2, and anti-derivatives in calculus. Anytime you are using an inverse function or inverse operation you want to "go on yellow alert" and pay close attention to domain issues.
     
  6. Oct 2, 2009 #5
    Re: a strange solution ??????

    Exactly =-) Even for something as common-place as division.... remember division doesn't *always* undo multiplication. 2 * 0 / 0 is NOT 2!!
     
  7. Oct 3, 2009 #6

    HallsofIvy

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    Re: a strange solution ??????

    I once had a student who complained bitterly that I marked his answer wrong on a test just because he used the "method he learned in high school" rather than the method I taught him. I don't think I was ever able to convince him that I marked him wrong because his answer did not satisfy the equation!
     
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