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Math Help - Easy quadratic solving

  1. #1
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    Easy quadratic solving

    If the roots of the quadratic equation (a b)x + (b c)x + (c a) = 0 are equal, prove that 2a = b + c.

    well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

    so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

    we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by ice_syncer View Post
    we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help
    Notice that (a-b)\cdot 1^2+(b-c)\cdot 1+c-a =  a-b+b-c+c-a=0 hence one is a root of the polynomial. That's why one has -\frac{B}{2A}=1.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ice_syncer View Post
    If the roots of the quadratic equation (a b)x + (b c)x + (c a) = 0 are equal, prove that 2a = b + c.

    well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

    so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

    we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help
    You could look at the discriminant of the original equaion and set it equal to 0. (That's the condition for two equal roots.)

    -Dan
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    Quote Originally Posted by ice_syncer View Post
    If the roots of the quadratic equation (a – b)x + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

    well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

    so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

    we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help
    Method 1: (School Method,Straightforward)

    If the roots are equal, then the discriminant is 0.

    So

    (b-c)^2 = 4(a-b)(c-a)

    b^2 + c^2 -2bc = 4(ac - a^2 -bc + ab)

    4a^2 + b^2 + c^2 + 2bc - 4ac - 4ab = 0

    (2a -b-c)^2 = 0

    2a -b-c = 0

    2a = b+c

    Method 2: (Simpler, Elegant)

    We see that 1 is a root of the equation since (a-b)(1)^2 + (b-c)1+(c-a) = 0

    So if the roots are equal, the other root also must be 1. This means the product of the roots is 1. Thus \frac{c-a}{a-b} = 1 \Rightarrow c-a = a-b \Rightarrow 2a = b+c
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  5. #5
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    Quote Originally Posted by flyingsquirrel View Post
    Hi


    Notice that (a-b)\cdot 1^2+(b-c)\cdot 1+c-a =  a-b+b-c+c-a=0 hence one is a root of the polynomial. That's why one has -\frac{B}{2A}=1.
    But it is a quadratic equation, so it can only have 2 roots at a time and those are the ones I found///
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by ice_syncer View Post
    But it is a quadratic equation, so it can only have 2 roots at a time and those are the ones I found///
    The polynomial still has two roots : -\frac{B}{2A}=1
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  7. #7
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    Quote Originally Posted by flyingsquirrel View Post
    The polynomial still has two roots : -\frac{B}{2A}=1
    yes yes I got itbefore only thanks anyywa
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