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Math Help - Solving for a surd

  1. #1
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    Solving for a surd

    (1+x)/x = sqrt3

    I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

    Any help please?
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  2. #2
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    Quote Originally Posted by 200001 View Post
    (1+x)/x = sqrt3

    I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

    Any help please?
    1+x = x \sqrt{3}

    1 = x\sqrt{3} - x

    1 = x(\sqrt{3} - 1)

    \frac{1}{\sqrt{3}-1} = x
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  3. #3
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    Hello 200001
    Quote Originally Posted by 200001 View Post
    (1+x)/x = sqrt3

    I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

    Any help please?
    You're right so far. Now solve this as you would any other equation: get the x's on one side and everything else on the other.
    1+x = x\sqrt3

    \Rightarrow 1 = x\sqrt3 -x
    Factorise:
    \Rightarrow 1 = x(\sqrt3-1)
    Divide:
    \Rightarrow x = \frac{1}{\sqrt3 -1}
    You can leave the answer like this, but it's neater if you 'rationalise the denominator':
    \Rightarrow x = \frac{1}{\sqrt3 -1}\color{red}\times \frac{\sqrt3+1}{\sqrt3+1}
    = \frac{\sqrt3+1}{3-1}

    =\frac{\sqrt3+1}{2}
    Grandad
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  4. #4
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    Both of you, thanks.....it was the final step that confused me
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  5. #5
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    Quote Originally Posted by 200001 View Post
    (1+x)/x = sqrt3

    I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

    Any help please?
    Treat \sqrt{3} like any other constant such as 4, 1 and -2.6

    The only real difference with surds is that you should never use a decimal approximation unless explicitly told to and you're expected to Rationalise the Denominator which means ensuring no surds are left on the bottom.

    To do this make use of the difference of two squares a^2-b^2=(a+b)(a-b)

    \frac{1}{\sqrt{3} -1} \times \frac{\sqrt3 +1}{\sqrt3 +1} = \frac{\sqrt3 +1}{(\sqrt3 -1)(\sqrt3 + 1)} = \frac{\sqrt3 +1}{3-1} = \frac{\sqrt3 +1}{2}
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