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Math Help - Surds problem

  1. #1
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    Surds problem

    This is a question from a local past paper. I am either having a mental breakdown or there is a problem in the question. As usual I need a sanity check please. I don't necessarily need the answer, but if you can tell me that it is solveable and there is not a mistake, then that is good.

    Using surds, find the values of a and b if \frac{\sqrt{3}+a}{\sqrt{3}+b}=\sqrt{3}-1
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  2. #2
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    Re: Surds problem

    Quote Originally Posted by grillage View Post
    This is a question from a local past paper. I am either having a mental breakdown or there is a problem in the question. As usual I need a sanity check please. I don't necessarily need the answer, but if you can tell me that it is solveable and there is not a mistake, then that is good.

    Using surds, find the values of a and b if \frac{\sqrt{3}+a}{\sqrt{3}+b}=\sqrt{3}-1
    a=-1 ~\&~ b=1-\sqrt{3} works.
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  3. #3
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    Re: Surds problem

    Hello, grillage!

    \text{Find the values of }a\text{ and }b\text{ if: }\:\frac{\sqrt{3}+a}{\sqrt{3}+b}\:=\:\sqrt{3}-1

    We have: . \sqrt{3} + a \;=\;(\sqrt{3}-1)(\sqrt{3}+b)

    . . . . . . . . \sqrt{3}+a \;=\;3 + b\sqrt{3} - \sqrt{3} - b

    n . . . a + b - b\sqrt{3} \;=\;3 - 2\sqrt{3}

    Equate rational and irreational components: . \begin{Bmatrix}a + b \:=\:3 \\ -b\sqrt{3} \:=\:-2\sqrt{3}\end{Bmatrix}
    Solve the system: . \begin{Bmatrix}a \:=\:1 \\ b \:=\:2\end{Bmatrix}
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Surds problem

    Multiply through by the denominator to get:

     \sqrt 3 +a = 3 - \sqrt 3 +b(\sqrt 3 -1)

    and rearrange to get b in terms of a:

     b = \frac {a -3 + 2 \sqrt 3 }{\sqrt 3 - 1}

    Now to pretty it up you can multiply through by  ( \sqrt 3 + 1)/(\sqrt 3 + 1) to get rid of the square root in the denominator. This gives an equation for 'b' as a function of 'a'. You should find for example that if 'a' = 1 then 'b' = 2, and if 'a' = -1 then b = 1-sqrt(3).
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