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Thread: what is happening here?

  1. #21
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    Quote Originally Posted by Dr.Peterson View Post
    Five is right. But along the way, two mistakes are made that cancel one another out. Getting the right answer in an invalid way doesn't count as correct.
    Thanks Doc. I simply did not look at the steps he took...
    I'm a man of few words...but I use 'em often!!

  2. #22
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    Quote Originally Posted by allegansveritatem View Post
    I'm sorry but this needs to be unpacked a little so a novice like me can make it out.
    Look at my comment and then at jomo's that immediately follows.

    If your first step is generally correct, then you do not need to change the sign when moving an addend from one side of an equation to the other.

    Is that generally correct? Let's experiment.

    [tex]3 + 7 = 10 \implies 7 = 10 + 3 \implies 7 = 13![/tex]

    When you got to the step where a square root equals a negative number, you should have known that you had made an error because square roots are non-negative by definition.

    Dr. P has suggested that you made two mistakes that offset each other. I am not 100% sure I agree with that despite my great respect for him. Your negative square root was not a separate mathematical mistake, but rather a logical consequence of your first mistake. Of course it should have warned you that something was very wrong with your logic, and failure to heed a warning is a mistake. But failing to heed a warning is not a type of mistake that offsets logical mistakes.

    Your question about why you nevertheless got the correct answer is an excellent one. It frequently happens that squaring an equation introduces spurious answers. You are normally told to check for such spurious answers after solving an equation by squaring. Here the squaring luckily introduced the "correct" answer. You should not count on such luck.

  3. #23
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    To make up for my previous trespass!!!

    √(m+4) + √(m-4) = 4

    square both sides:
    m+4 + 2√(m+4)√(m-4) + m-4 = 16

    simplify:
    2√(m+4)√(m-4) = 16 - 2m
    √(m^2 - 16) = 8 - m

    square both sides:
    m^2 - 16 = 64 - 16m + m^2

    simplify:
    16m = 64 + 16
    16m = 80
    m = 5

    EDIT: .....and I will recite the Hoooooly Rosary twice, skipping no beads
    Last edited by Denis; 01-11-2019 at 03:33 PM.
    I'm a man of few words...but I use 'em often!!

  4. #24
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    Quote Originally Posted by Denis View Post
    To make up for my previous trespass!!!

    √(m+4) + √(m-4) = 4

    square both sides:
    m+4 + 2√(m+4)√(m-4) + m-4 = 16

    simplify:
    2√(m+4)√(m-4) = 16 - 2m
    √(m^2 - 16) = 8 - m

    square both sides:
    m^2 - 16 = 64 - 16m + m^2

    simplify:
    16m = 64 + 16
    16m = 80
    m = 5
    A very nice demonstration of the fact that, although teachers say to isolate one radical before squaring, that's not essential; it just usually makes the work easier. You can do it either way, as long as you do each step correctly. (This way, you had to square binomials twice, which many students either do wrong, or find difficult).

    Of course, at the end you had to check that the solution isn't extraneous, though you didn't show it ...

  5. #25
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    Thanks Doc...see my edit!!
    I'm a man of few words...but I use 'em often!!

  6. #26
    Quote Originally Posted by JeffM View Post
    Look at my comment and then at jomo's that immediately follows.

    If your first step is generally correct, then you do not need to change the sign when moving an addend from one side of an equation to the other.

    Is that generally correct? Let's experiment.

    [tex]3 + 7 = 10 \implies 7 = 10 + 3 \implies 7 = 13![/tex]

    When you got to the step where a square root equals a negative number, you should have known that you had made an error because square roots are non-negative by definition.

    Dr. P has suggested that you made two mistakes that offset each other. I am not 100% sure I agree with that despite my great respect for him. Your negative square root was not a separate mathematical mistake, but rather a logical consequence of your first mistake. Of course it should have warned you that something was very wrong with your logic, and failure to heed a warning is a mistake. But failing to heed a warning is not a type of mistake that offsets logical mistakes.

    Your question about why you nevertheless got the correct answer is an excellent one. It frequently happens that squaring an equation introduces spurious answers. You are normally told to check for such spurious answers after solving an equation by squaring. Here the squaring luckily introduced the "correct" answer. You should not count on such luck.
    The second sentence of the above is certainly NOT true, right? I mean, when is it ever true?
    I strode past that part--the part where the negative number appeared on one side and a lonely radical on the other--like a giant racing to the finish line. When I say "like a giant" you should see in imagination a giant rhinoceros in mid charge.

  7. #27
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    Quote Originally Posted by Denis View Post
    To make up for my previous trespass!!!

    √(m+4) + √(m-4) = 4

    square both sides:
    m+4 + 2√(m+4)√(m-4) + m-4 = 16

    simplify:
    2√(m+4)√(m-4) = 16 - 2m
    √(m^2 - 16) = 8 - m

    square both sides:
    m^2 - 16 = 64 - 16m + m^2

    simplify:
    16m = 64 + 16
    16m = 80
    m = 5

    EDIT: .....and I will recite the Hoooooly Rosary twice, skipping no beads
    No check, no credit. Go to the corner, again.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  8. #28
    Quote Originally Posted by Dr.Peterson View Post
    A very nice demonstration of the fact that, although teachers say to isolate one radical before squaring, that's not essential; it just usually makes the work easier. You can do it either way, as long as you do each step correctly. (This way, you had to square binomials twice, which many students either do wrong, or find difficult).

    Of course, at the end you had to check that the solution isn't extraneous, though you didn't show it ...
    Well, I checked in the sense that it was easy to see that 5 was the obvious and only answer. But I thought that when an equation has only one answer, like this one, there can be no extraneous root to check, no?

  9. #29
    Finally (this time I mean it):
    m42.jpg

  10. #30
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    Quote Originally Posted by Jomo View Post
    No check, no credit. Go to the corner, again.
    That's where I recited my 2 Hooooly rosaries Father Jomo!!
    I'm a man of few words...but I use 'em often!!

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