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I can't solve a simple equation. Help please.

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intabits:
I haven't routinely juggled algebraic expressions since school over 45 years ago. I do occasionally find myself needing to do so again, and usually with success, but this time I'm stumped.

My program's calculation brings me to point where I'm trying to calculate a value from two others already known, and the equation boils down to something of the form:-

a = (1/x) - 1/(x+b)         (1)

I can't solve this (seemingly simple) equation for x.
a & b are not constants, they're calculated in preceding steps. Since the equation (1) has only one unknown, I expect that there must be some solution of the form:-

x = f(a,b)   i.e. some function of a & b

Then my program (dealing with resistors in parallel and series) can plug in a & b to get a value for x, and continue on its merry way.

But I cannot reduce the two occurrences of x in eqn(1) to just one. I seem only able to make it worse, such as:-

x+b = ax(x+b) + x               with four occurrences of x.

Can someone please provide me with the right hand side of:-

x = ?

(Or explain why it is not possible)
Thanks in advance!



mawyatt:
Classic quadric equation of x^2 + bx - b/a.

Best,

Spike101:
You can always cheat... :)

https://www.wolframalpha.com/input/?i=solve+a%3D%281%2Fx%29-%281%2F%28x%2Bb%29%29+for+x

dmills:
a = (1/x) - 1/(x+b)         (1)
Multiply by (x+b)
a(x+b) = (x+b)/x - (x+b)/(x+b) = x/x + b/x - (x+b)/(x+b) = 1+b/x  - 1 = b/x
Multiply by x
ax(x+b) = b
subtract b
ax^2 + abx -b = 0
Divide thru by a.
x^2 + bx - b/a

Solve as a quadratic by the usual means (Note there are two roots, and values of a and b that will result in x being complex).

Someone check my working, I may have screwed up a sign somewhere or something, it has been a while.

rf-messkopf:

--- Quote from: mawyatt on February 02, 2021, 01:01:56 pm ---Classic quadric equation of x^2 + bx - b/a.

--- End quote ---

Correct. More precisely, rearranging yields the equation x^2 + bx - b/a = 0. This has two solutions:
\[x_{1,2}=-\frac b2\pm\sqrt{\frac{b^2}{4}+\frac ba}.\]

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