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WEMG · Feb 22, 2010

Solve:
[(3m+1)/(3m-1)]-[(3m-1)/(3m+1)]=2

Answer: m=(1/3)(1+2^1/2) or m= (1/3)(1-2^1/2)

I need solution

Drongoski · Feb 22, 2010

WEMG · Feb 22, 2010

Drongoski said:
$\100dpi \frac{3m+1}{3m-1}-\frac{3m-1}{3m+1}\ =\ 2\ \ \Rightarrow \frac{(3m+1)^{2}-(3m-1)^{2}}{(3m-1)\\ \\ (3m+1)}\ =\ 2\\ \\ \Rightarrow (3m+1)^{2}\ -\ (3m-1)^{2}\ =\ 2([3m]^{2}-1^{2})\\ \\ \Rightarrow \ \ (3m+1 + 3m-1)(3m+1 - 3m+1)\ =\ 2(9x^{2}-1)\\ \\\Rightarrow \ \ 6m\times 2 \ =\ 2(9m^{2}-1)\ \ \Rightarrow 9m^{2}-6m-1\ =\ 0\ \ (quadratic\ eqn\ in\ m)\\ \\ \therefore \ \ m\ =\ \frac{-(-6)\ \pm \sqrt{(-6)^{2}-4\cdot 9\cdot [-1]})}{2\cdot 9}\ \ =\ \ \frac{1\ \pm \ \sqrt{2}}{3}$

Ahh, I accidently expanded when I should have factorise difference of 2 squares.
Thnks!

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WEMG

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Drongoski

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