A strange solution ?

  • Thread starter vikcool812
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  • #1
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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?????????
 

Answers and Replies

  • #2
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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.
 
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  • #3
HallsofIvy
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Or: [itex]\sqrt{x^2}= |x|[/itex], not just "x".
 
  • #4
jambaugh
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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.
 
  • #5
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Anytime you are using an inverse function or inverse operation you want to "go on yellow alert" and pay close attention to domain issues.
Exactly =-) Even for something as common-place as division.... remember division doesn't *always* undo multiplication. 2 * 0 / 0 is NOT 2!!
 
  • #6
HallsofIvy
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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.
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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