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Math Help - where did I go wrong in this equation?

  1. #1
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    where did I go wrong in this equation?

    2x-12=-2(x^2-3x)^(.5)

    1) divide by -2 by the associative property of mult.
    2) that makes -x+6=(x^2-3x)^(.5), then square both sides
    3) that makes, x^2-12x+36=(x^2-3x)
    4) then, subtract x^2 and x^2, and divide -3 and -12 to make 4 and subtract 36.
    5) that makes 4x=x-36, so subtract x to make 3x=-36 and divide by 3 to make x=-12 but the answer is x=4. What happened?

    It seems to me like all the properties of numbers and order of operations were followed
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  2. #2
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    x^2-12x+36=x^2-3x

    -12x+36=-3x

    Add 12x to both sides

    36=15x

    x= \frac{36}{15} = \frac{12}{5}
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  3. #3
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    0) Have you thought about the Domain? For Real Numbers, everything under the radical must manage to be non-negative. So, x <= 0 or x >= 3
    1) Why is that the associateive property of multiplication. Think about multiplicative inverse.
    1) Have you thought about the Domain? For Real Numbers, 6-x >= 0. So, x < 6
    2) This is why we think about the Domain BEFORE we do things like this.
    4) Whoops. This is in error. Also, 36 / (-3) = -12, giving 4x - 12 = x

    Be about a 1000 times more careful. Give it another go.
    Last edited by TKHunny; April 11th 2011 at 06:39 PM.
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  4. #4
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    Quote Originally Posted by bournouli View Post
    2x-12=-2(x^2-3x)^(.5)

    1) divide by -2 by the associative property of mult.
    2) that makes -x+6=(x^2-3x)^(.5), then square both sides
    3) that makes, x^2-12x+36=(x^2-3x)
    4) then, subtract x^2 and x^2, and divide -3 and -12 to make 4 and subtract 36.
    After subtracting x^2 from both sides you get -12x+ 36= -3x. Dividing both sides by -3 gived you 4x- 12= x. You did not divide 36 by -3.


    5) that makes 4x=x-36, so subtract x to make 3x=-36 and divide by 3 to make x=-12 but the answer is x=4. What happened?

    It seems to me like all the properties of numbers and order of operations were followed
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pickslides View Post
    x^2-12x+36=x^2-3x

    -12x+36=-3x

    Add 12x to both sides

    36=15x
    You lost a negative sign in there. The equation should be 36 = 9x.

    -Dan
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    After subtracting x^2 from both sides you get -12x+ 36= -3x. Dividing both sides by -3 gived you 4x- 12= x. You did not divide 36 by -3.
    That's the rub -you don't have to divide both sides just both terms:

    -12x+36=-3x, by a property of mult. (a*b)=(a)*b, so only -12 and -3 enter the relation of division (and division is just a type of mult.), so finally one can conclude that 4x+36=x. That then leads into 4x-x=3x and 36 is subtracted from both sides to make -36=3x which means x=-12
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bournouli View Post
    That's the rub -you don't have to divide both sides just both terms:

    -12x+36=-3x, by a property of mult. (a*b)=(a)*b, so only -12 and -3 enter the relation of division (and division is just a type of mult.), so finally one can conclude that 4x+36=x
    Asked, and answered! As already stated you have to divide both terms:
    \displaystyle -12x + 36 = -3x

    \displaystyle \frac{-12x + 36}{-3} = x

    \displaystyle \frac{-12x}{-3} + \frac{36}{-3} = x

    4x - 12 = x

    Consider the eample, doing it your way:
    \displaystyle \frac{9}{3} = \frac{3 + 6}{3} = \frac{3}{3} + 6

    where I have divided only the leading term as you are suggesting. But
    \displaystyle \frac{9}{3} = ~...~ = \frac{3}{3} + 6 = 1 + 6 = 7

    an obviously incorrect result.

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Asked, and answered! As already stated you have to divide both terms:
    \displaystyle -12x + 36 = -3x

    \displaystyle \frac{-12x + 36}{-3} = x

    \displaystyle \frac{-12x}{-3} + \frac{36}{-3} = x

    4x - 12 = x

    Consider the eample, doing it your way:
    \displaystyle \frac{9}{3} = \frac{3 + 6}{3} = \frac{3}{3} + 6

    where I have divided only the leading term as you are suggesting. But
    \displaystyle \frac{9}{3} = ~...~ = \frac{3}{3} + 6 = 1 + 6 = 7

    an obviously incorrect result.

    -Dan
    Why can't 7/3=9/3 except for the obvious contradiction? Here's how I would derive that result: (3)(1/3)+(6)(1/3), by Associative law of Addition and Mult: ((3)(1/3)+(6)) (1/3), PEMDAS: (1+6=7)(1/3), so the answer is 7/3.

    Please dumb down your response -I'm not feeling too smart after writing that.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bournouli View Post
    Why can't 7/3=9/3 except for the obvious contradiction?
    If it gives a contradiction then it isn't correct!

    Quote Originally Posted by bournouli View Post
    Here's how I would derive that result: (3)(1/3)+(6)(1/3), by Associative law of Addition and Mult: ((3)(1/3)+(6)) (1/3), PEMDAS: (1+6=7)(1/3), so the answer is 7/3.

    Please dumb down your response -I'm not feeling too smart after writing that.
    You are using your laws wrong. Here, let me restate them:
    Associative Laws
    (a + b) + c = a + (b + c)
    (ab)c = a(bc)

    Commutative Laws
    a + b = b + a
    ab = ba

    Distributive Laws
    a(b + c) = ab + ac
    (a + b)c = ac + bc

    What you are trying to use is the reverse of the Distributive Law, ac + bc = (a + b)c, but you are writing ac + bc = (ac + b)c. This is obviously not correct. You have to factor the c (in your case 1/3) from both terms.

    That is to say
    \displaystyle 3 \cdot \frac{1}{3} + 6 \cdot \frac{1}{3} = (3 + 6) \frac{1}{3}

    not
    \displaystyle 3 \cdot \frac{1}{3} + 6 \cdot \frac{1}{3} = \left ( 3 \cdot \frac{1}{3}  + 6 \right )  \frac{1}{3}

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark View Post
    If it gives a contradiction then it isn't correct!


    You are using your laws wrong. Here, let me restate them:
    Associative Laws
    (a + b) + c = a + (b + c)
    (ab)c = a(bc)

    Commutative Laws
    a + b = b + a
    ab = ba

    Distributive Laws
    a(b + c) = ab + ac
    (a + b)c = ac + bc

    What you are trying to use is the reverse of the Distributive Law, ac + bc = (a + b)c, but you are writing ac + bc = (ac + b)c. This is obviously not correct. You have to factor the c (in your case 1/3) from both terms.

    That is to say
    \displaystyle 3 \cdot \frac{1}{3} + 6 \cdot \frac{1}{3} = (3 + 6) \frac{1}{3}

    not
    \displaystyle 3 \cdot \frac{1}{3} + 6 \cdot \frac{1}{3} = \left ( 3 \cdot \frac{1}{3}  + 6 \right )  \frac{1}{3}

    -Dan
    That helps but for just one question -Isn't the Distributive Property just an instance of the Associative Property. The latter says If you have addition or mult. then you can arrange the terms in any way you want: a*b=b*a. In my example I have ac+bc. That is, I have an instance of mult. (ac and bc). So I can arrange the terms in any way I like, so ac+bc becomes ac+b(c) and I also have an instance of addition (ac+b) so I will arrange terms in any way: (ac+b)(c). So that is possible (it's not I'm just giving an argument). What would be the right definition of the properties of numbers?
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  11. #11
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    2x - 12 = -2(x² - 3x)^½
    6 - x = (x² - 3x)^½
    (6 - x)² = ((x² - 3x)^½)²
    x² - 12x + 36 = x² - 3x
    36 = 12x - 3x = 9x
    9x = 36
    ∴ x = 36/9 = 4
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  12. #12
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    Quote Originally Posted by bournouli View Post
    That helps but for just one question -Isn't the Distributive Property just an instance of the Associative Property. The latter says If you have addition or mult. then you can arrange the terms in any way you want: a*b=b*a. In my example I have ac+bc. That is, I have an instance of mult. (ac and bc). So I can arrange the terms in any way I like, so ac+bc becomes ac+b(c) and I also have an instance of addition (ac+b) so I will arrange terms in any way: (ac+b)(c). So that is possible (it's not I'm just giving an argument). What would be the right definition of the properties of numbers?
    Well, among other things, the Associative Law only works with three terms or factors, not two. The Associative Law cannot be applied to ac + bc, only the Distributave Law.

    -Dan

    Edit: You have stated ac + bc and ac + b(c). Order of operations dictates that multiplication comes before addition, so you cannot say this is ac + b times c. Look at it this way. (ac + b)c = ac^2 + bc not ac + bc.
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  13. #13
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    Quote Originally Posted by bournouli View Post
    That's the rub -you don't have to divide both sides just both terms:

    -12x+36=-3x, by a property of mult. (a*b)=(a)*b, so only -12 and -3 enter the relation of division (and division is just a type of mult.), so finally one can conclude that 4x+36=x. That then leads into 4x-x=3x and 36 is subtracted from both sides to make -36=3x which means x=-12
    Yes, that is, indeed, "the rub"- you are completely wrong. You cannot divide just chosen parts of an equation by a number, you must divide every term in an equation by the divisor.
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