(Sturm-Liouville) I'm having trouble understanding why the root of a given characteristic equation is the value that it is. More information posted below.

Question

mendicant_bias · Answer

Take it as given that the characteristic equation of a given SLDE is: 
$$\large{9r^2-2r+\lambda=0}$$Wouldn't the root of this be found by using the quadratic formula? I'm having a lot of trouble understanding why their result using it is so different from mine, unless there's something different about the formula specifically with char. eqns that I'm forgetting.

mendicant_bias · Answer

E.g., I would try to find the root of the characteristic equation like this: $$r = \frac{-(-2)\pm\sqrt{(-2)^2-4(9)(\lambda)}}{2(9)}$$
But that isn't correct. What am I doing wrong? @SithsAndGiggles

mendicant_bias · Answer

(This is coming from PDF page 2, problem 2, of the following document): http://www4.ncsu.edu/~lkn/MA401/SolutionsToTests/FinalExam.pdf I just don't understand how they got their root.

mendicant_bias · Answer

@AllTehMaffs

mendicant_bias · Answer

I mean, this is a final exam solutions set, so I highly doubt it's a series of horribly egregious errors.

anonymous · Answer

They simplified it a whole bunch by bringing the 2 on bottom up top

$$ r = \frac{\frac{1}{2}2 \pm \frac{1}{2}\sqrt{4-4*9*\lambda}}{9} $$

^_^

mendicant_bias · Answer

Alright, cool, that's a lot less worrying and makes a ton more sense.

anonymous · Answer

w00t