# How to factor a quadratic function based on variables not numbers?

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:

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

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.

• So is the equation correct if written as I've put an addendum in my question?
– mike
Jan 7, 2020 at 5:33
• Aside from the stray minus sign, that is correct. You have $-*$ between the two factors. Jan 7, 2020 at 5:35
• Yeah Ross that was a commenting glitch. I moved it to my answer. Is that now correct? Thanks again.
– mike
Jan 7, 2020 at 5:36

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

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

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

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.