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If I have the following equation where $C$, $L$, and $R$ are constants, how do I solve for $a$ and $b$?

$$s^2 + \frac{s}{2CR}+ \frac{1}{CL} = (s+a)(s+b)$$

I can put the quadratic ($s^2+s/(2CR)+1/(CL)$) into wolfram alpha and ask it to solve the equation assuming it equals zero which gives me:

$$s = -\frac{\sqrt{L (L - 16 C R^2)} + L}{4 C L R}$$ $$s = \frac{\sqrt{L (L - 16 C R^2)} - L}{4 C L R}$$

Does that help me in some way?

I am unsure of how to do this and can't find any guide online that seems to apply. Thanks for any help.

Addendum - I'm told those are the two solutions for $a$ and $b$. So is this correct:

$s^2 + \frac{s}{2CR}+ \frac{1}{CL} = (s-(sqrt(L (L - 16 C R^2)) + L)/(4 C L R))*(s+(sqrt(L (L - 16 C R^2)) - L)/(4 C L R))$

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3 Answers 3

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Yes, these are the two roots, $-a$ and $-b$. If you have values for $L,C,R$ you can plug them in. Alpha has just applied the quadratic formula for you.

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  • $\begingroup$ So is the equation correct if written as I've put an addendum in my question? $\endgroup$
    – mike
    Jan 7, 2020 at 5:33
  • $\begingroup$ Aside from the stray minus sign, that is correct. You have $-*$ between the two factors. $\endgroup$ Jan 7, 2020 at 5:35
  • $\begingroup$ Yeah Ross that was a commenting glitch. I moved it to my answer. Is that now correct? Thanks again. $\endgroup$
    – mike
    Jan 7, 2020 at 5:36
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$s^2 + (a+b)s + ab = s^2 + \frac {1}{2CR} s + \frac {1}{CL}$

If the LHS equals the RHS for all $s$ then $a+b = \frac{1}{2CR}$ and $ab = \frac{1}{CL}$

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$$ s^2 + \frac{s}{2CR}+ \frac{1}{CL} = (s+a)(s+b) $$

I take it you want both $a$ and $b$ as functions of $s,C,R,L.$

$$ s^2 + \frac{s}{2CR}+ \frac{1}{CL} = (s+a)(s+b) = s^2 + s(a+b) + ab $$

$$ \frac{s}{2CR}+ \frac{1}{CL} = s(a+b) + ab $$ $$ \frac s {2CR} - s(a+b) = \frac 1 {CL} + ab $$

$$ s\left( \frac 1 {2CR} -(a+b) \right) = \frac 1 {CL} + ab $$

As $s$ changes, the right side of this does not change. Therfore the left side does not change. But that can happen only if

$$ \frac 1 {2CR} - (a+b) = 0. $$

That tells you what $a+b$ is. PROVIDED the idea is that this should be just one solution that holds no matter what number $s$ is. (If that's not what is meant, then disregard all of this.

Now you've got the left side equal to $0,$ so $$ 0=\frac 1 {CL} + ab. $$

That tells you what $ab$ is.

If you know $a+b$ and you know $ab,$ how do you find $a$ and $b.$

One way is this: You have $a+b= \text{something}.$ So $b=(\text{something} - a).$

And $ab = \text{something else}.$ So $a\big( \text{something} - a\big) = \text{something else}.$

That equation is quadratic in $a,$ so you can use the usual formula.

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