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Thread: Whats wrong with this mathematically?

  1. #1

    Default Whats wrong with this mathematically?

    When solving a quadratic equation, such as this, we need to factorize first:

    x^2+6x+8 = 0

    => (x+4)(x+2) = 0

    So x = -4 or x = -2

    Why can't we do x(x+6) = -8?
    And so x = -8 or x = -6 (obviously wrong)

    Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with...

  2. #2
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    Oh my... my maths is e kind of straight F9s...


    *this brings back nightmares....*

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    Quote Originally Posted by David
    When solving a quadratic equation, such as this, we need to factorize first:

    x^2+6x+8 = 0

    => (x+4)(x+2) = 0

    So x = -4 or x = -2

    Why can't we do x(x+6) = -8?
    And so x = -8 or x = -6 (obviously wrong)

    Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with...
    second equation is correct

    in the second equation, x is still = -4 or -2

    if x = -8

    x(x+6) = -8(-8+6) = 16

    if x = -6

    x(x+6) = -6(-6+6) = 0

  4. #4

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    think the RHS has to be 0 because when 2 terms multiply with each other, (x+4)(x+2), the only way the result is 0 is that one of the term is zero. this way you can bring out each term and suppose that it is 0. that's why the 'or' in the answer.

    however for x(x+6) = -8, x could be anything like -2, -1, -8... anything.. and (x+6) could be 4, 8, 1... anything, in accordance to the assumed value of x. You probably can find the answer too this way but i guess it's an exhausive search kind of thing. Hope this helps.

  5. #5

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    hey,
    from
    x(x+a)=0
    you can get x=0 or x=-a;
    but there's no direct conclusion to be made if the right hand side is not ZERO...

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    Quote Originally Posted by David
    When solving a quadratic equation, such as this, we need to factorize first:

    x^2+6x+8 = 0

    => (x+4)(x+2) = 0

    So x = -4 or x = -2

    Why can't we do x(x+6) = -8?
    And so x = -8 or x = -6 (obviously wrong)

    Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with...
    Multiply any number with zero and the outcome is zero. So if A * B = 0 then one or both of them must be zero.

    As such, when a quadratic equation can be factored down to (x+4)(x+2) = 0, then either X+4 = 0 or X+2 = 0. So there are 2 solutions, X=-4 or X=-2.

    If A * B = C and C<>0, then A=C only when B=1, or B=C only when A=1.

    So if you re-express the euation in the form x(x+6) = -8, then it does not imply that X=-8 or X+6 = -8.
    As complexity rises, precise statements lose meaning and meaningful statements lose precision.

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    "ab = 0" iff "a = 0 or b = 0" for all real a, b.

    i.e. if a product for two numbers is zero, then at least one of them must be zero. Likewise, like Roy mentioned, any product with a zero gives you zero.

    Same reason cannot hold for a product that is non-zero because there are infinite ways to give a non-zero product involving real numbers.

    There is a finite number of ways to give you a non-zero product IF you are restricted to only integer solutions though:

    For example, ab = -8 iff
    a = -8, b = 1 or
    a = -4, b = 2 or
    a = -2, b = 4 or
    a = -1, b = 8 or
    a = 1, b = -8 or
    a = 2, b = -4 or
    a = 4, b = -2 or
    a = 8, b = -1.

    If none of these combinations solve the given equation then we can say that the equation has no integer solution.

  8. #8

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    Wow thanks guys... quite a read!!!! Yup, makes sense..

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    Quote Originally Posted by David

    Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with...
    This is because you're finding the roots of the equation, ie The x-intercepts. To find this, you'll have to equate y=o, and hence from the equation "y=ax^2+bx+c", you'll get "ax^2+bx+c=0" and you solve for x. Hope this helps =)
    eat. drink. shoot

  10. #10

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    Quote Originally Posted by +evenstar
    This is because you're finding the roots of the equation, ie The x-intercepts. To find this, you'll have to equate y=o, and hence from the equation "y=ax^2+bx+c", you'll get "ax^2+bx+c=0" and you solve for x. Hope this helps =)
    Thanks evenstar... But I dun think that's the reason... I'll go with the ones given by the rest of the people above.

    Anyway, thanks.

  11. #11

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    Quote Originally Posted by David
    Thanks evenstar... But I dun think that's the reason... I'll go with the ones given by the rest of the people above.

    Anyway, thanks.
    I think it was a sound reasoning.

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    Quote Originally Posted by asturias105
    I think it was a sound reasoning.
    Err...it's how we do math in Seconday School. Others have higher educational level than me, hence they may have better knowledge than me...
    eat. drink. shoot

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    many many years didn't go maths liao

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    go draw out the graph of the equation you stated.

    draw a line y = 8. you will find that this line will intersect at x=-6 and x=0.

    as we want the "roots", the interesection of the curve with the x-axis. Naturally, we have let y=0.

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    Default Quadratic....

    Quote Originally Posted by David
    When solving a quadratic equation, such as this, we need to factorize first:

    x^2+6x+8 = 0

    => (x+4)(x+2) = 0

    So x = -4 or x = -2

    Why can't we do x(x+6) = -8?
    And so x = -8 or x = -6 (obviously wrong)

    Shucks, learnt this in sec sch but cant remeber the reason... only know it's illegal unless the RHS is a zero to begin with...
    Solution where the curve cuts the x axis....
    the discriminant...b^2-4ac tells you what type of solution you gonna get....
    b^2-4ac > 0 two roots real and different -> this is yours x = -4 or x = -2
    b^2-4ac = 0 two roots real and equal
    b^2-4ac < 0 two roots which are maginary...


    rgds,
    sulhan

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    Quote Originally Posted by sulhan
    Solution where the curve cuts the x axis....
    the discriminant...b^2-4ac tells you what type of solution you gonna get....
    b^2-4ac > 0 two roots real and different -> this is yours x = -4 or x = -2
    b^2-4ac = 0 two roots real and equal
    b^2-4ac < 0 two roots which are maginary...


    rgds,
    sulhan
    Err...

    b^2 - 4ac > 0 means 2 real roots (ie graph intercepts x-axis at 2 points)
    b^2 - 4ac = 0 means 1 real root or 2 real roots which are equal (ie graph intercepts x axis at 1 point only)
    b^2 - 4ac < 0 means no real roots (ie graph does not intercept x-axis)
    eat. drink. shoot

  17. #17
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    I think David's question is, in essence, asking for the reason why
    "ab = 0" lead to "a=0 or b=0"
    but
    "ab = c (where c is non-zero)" cannot give a conclusion the same way, i.e. lead to "a = c or b = c" or something similar. Not asking why we need to equate to zero instead of other things.

    In fact, we CAN write an equation of the form x(x+6) = -8 and solve it like what slacker123 has described, i.e. draw the graphs of y = x(x+6) and y = -8 and see where they intersect. (But if we really wish to draw the graph, it makes more sense to draw y = x^2 + 6x + 8 and see where it intersects with y = 0, i.e. x-axis.)

    Mathematically they yield the same solutions because the graph of y1 = x^2 +6x + 8 = (x+2)(x+4) is actually the graph of y2 = x(x+6) shifted upwards by 8-units. i.e. y1 = (x+2)(x+4) = x^2 + 6x + 8 = x(x+6) + 8 = y2 + 8. Thus, the x-coordinate where graph of y1 intersects y = 0 is the same as where y2 intersects y = -8, i.e 8 units below. Hope this should explain eikin's and slacker123's ideas.

  18. #18

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    Quote Originally Posted by David
    Why can't we do x(x+6) = -8?
    And so x = -8 or x = -6 (obviously wrong)

    step 1 : "x(x+6) = -8"

    step 2: "hence x = -8 or x = -6"


    because from step 1 to step 2, the logic is lost already.
    i.e. step 2 is not a logical consequence of step 1. and how do we verify that from step 1 to step 2 is incorrect? simple. just subst x = -6 or x = -8 into x(x+6). u will never get -8

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