Originally Posted by

**TD!** Ok, so far so good! You know have m/(...) = 1.

A fraction is 1 if numerator and denominator are equal.

So you can easily solve this for m. We have:

$\displaystyle

m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}}

$

Now, this m doesn't seem "constant", but dependent of x, c and k!

But, where do we have to look for ortogonality? At their intersection!

Let's try to find something useful, equalling both equations:

$\displaystyle

\frac{1}{{x + c}} = m\left( {x + k} \right)^{\frac{1}{3}} \Leftrightarrow \frac{1}{m} = \left( {x + c} \right)\left( {x + k} \right)^{\frac{1}{3}} \Rightarrow \frac{1}{{m^2 }} = \left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}}

$

You recognise this? Filling in, in what we had for m earlier:

$\displaystyle

m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} \Leftrightarrow m = \frac{3}{{m^2 }} \Leftrightarrow m^3 = 3 \Leftrightarrow m = \sqrt[3]{3}

$