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    Name:  step question.JPG
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Size:  24.2 KBThe first part is fairly straightforward but I'm a bit stuck with the second part.
    I know that you have to consider the discriminant, but i cant seem to get anywhere and end up with a horrible Quartic Equation.
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    You shouldn't get a quartic. Post your workings do I can see what you have done?
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    How have you ended up with a quartic?


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    We're not going to magically see what you've done, you need to post your working.
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    Your method sounds right, try doing the algebra again.

    For reference:

    Spoiler:
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\dfrac{x}{x-a} + \dfrac{x}{x-b} = c+1

\Rightarrow x(x-b) + x(x-a) = (c+1)(x-a)(x-b)

\Rightarrow x^2 -bx +x^2-ax = (c+1)(x^2-ax-bx+ab)

\Rightarrow 2x^2 -bx-ax = cx^2 -cax-cbx+cab+x^2-ax-bx+ab

\Rightarrow x^2-cx^2+cax+cbx-cab-ab = 0 

\Rightarrow (1-c)x^2 + (ca+cb)x + (-cab-ab) = 0

\text{For exactly one real root, } (ca+cb)^2 - 4(1-c)(-cab-ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(1-c)(cab+ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(cab+ab-c^2ab-cab)=0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4ab - 4c^2ab = 0

\Rightarrow c^2(a^2-2ab+b^2) = -4ab

\Rightarrow c^2(a-b)^2 = -4ab

\Rightarrow c^2 = - \dfrac{4ab}{(a-b)^2} \text{, as required.}
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    (Original post by Forecast)
    Your method sounds right, try doing the algebra again.

    For reference:

    Spoiler:
    Show

    

\dfrac{x}{x-a} + \dfrac{x}{x-b} = c+1

\Rightarrow x(x-b) + x(x-a) = (c+1)(x-a)(x-b)

\Rightarrow x^2 -bx +x^2-ax = (c+1)(x^2-ax-bx+ab)

\Rightarrow 2x^2 -bx-ax = cx^2 -cax-cbx+cab+x^2-ax-bx+ab

\Rightarrow x^2-cx^2+cax+cbx-cab-ab = 0 

\Rightarrow (1-c)x^2 + (ca+cb)x + (-cab-ab) = 0

\text{For exactly one real root, } (ca+cb)^2 - 4(1-c)(-cab-ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(1-c)(cab+ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(cab+ab-c^2ab-cab)=0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4ab - 4c^2ab = 0

\Rightarrow c^2(a^2-2ab+b^2) = -4ab

\Rightarrow c^2(a-b)^2 = -4ab

\Rightarrow c^2 = - \dfrac{4ab}{(a-b)^2} \text{, as required.}
    They haven't listed a method, they've literally just suggested an approach, also listing the full solution is a bit counter productive


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    (Original post by Forecast)
    Your method sounds right, try doing the algebra again.

    For reference:

    Spoiler:
    Show

    

\dfrac{x}{x-a} + \dfrac{x}{x-b} = c+1

\Rightarrow x(x-b) + x(x-a) = (c+1)(x-a)(x-b)

\Rightarrow x^2 -bx +x^2-ax = (c+1)(x^2-ax-bx+ab)

\Rightarrow 2x^2 -bx-ax = cx^2 -cax-cbx+cab+x^2-ax-bx+ab

\Rightarrow x^2-cx^2+cax+cbx-cab-ab = 0 

\Rightarrow (1-c)x^2 + (ca+cb)x + (-cab-ab) = 0

\text{For exactly one real root, } (ca+cb)^2 - 4(1-c)(-cab-ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(1-c)(cab+ab) = 0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4(cab+ab-c^2ab-cab)=0

\Rightarrow c^2a^2 + 2c^2ab + c^2b^2 + 4ab - 4c^2ab = 0

\Rightarrow c^2(a^2-2ab+b^2) = -4ab

\Rightarrow c^2(a-b)^2 = -4ab

\Rightarrow c^2 = - \dfrac{4ab}{(a-b)^2} \text{, as required.}
    Unless im high Implications r wrong way as they said if c^2 then so word your argument carefully. You probably wouldn't get docked in this case I think.


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    (Original post by drandy76)
    They haven't listed a method, they've literally just suggested an approach, also listing the full solution is a bit counter productive


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    Op said considering discriminant


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    (Original post by physicsmaths)
    Op said considering discriminant


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    I guess in this case that's pretty much all you need to know tbf


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