How to solve symbolic function like this in R - r

Question

This should be simple for those with experience.
I want to solve an equation using R. I know you can solve
different linear/quadratic equations using Solve().
But I have something like this:
1/20 = 1/8 * (1/(12+x)) + 1/4*(1/(40+x)) + 3/4*(1/(50+x))
How can I solve x in this case? It can't be done by hand.
It gotta be some numeric methods involved to solve this like in TI83.
Is there a simple and quick way to do this in R without writing lines of codes?
Thank you!

Answer 1

As you say, there are indeed roots. First thing is to plot the function:
f <- function(x) {1/20 - 1/8 * (1/(12+x)) + 1/4*(1/(40+x)) + 3/4*(1/(50+x))}
x <- seq(-100,100)
par(mar=c(2,2,1,2)) # this just minimizes plot margins
plot(x,f(x), type="l")
abline(0,0,col="blue",lty=2)
So obviously, f(x) does cross 0, several times.
Next step is to estimate the crossings. One way to do this is to look for changes in sign:
x <- seq(-75,0,0.001)
y <- sign(f(x)) # vector of +1 or -1
plus.to.minus <- which(diff(y)<0) # diff(y)<0 when f crosses from (+) to (-)
minus.to.plus <- which(diff(y)>0) # diff(y)>0 when f crosses from (-) to (+)
# first two roots are (+) to (-); third is (-) to (+)
lower <- c(plus.to.minus[1:2],minus.to.plus[3])
roots <- sapply(lower,function(i)uniroot(f,interval=c(x[i],x[i+1]))$root)
lapply(roots,function(x) points(roots,c(0,0,0),col="red",pch=16))
roots
# [1] -67.38961 -41.72593 -10.38446
This code attempts to find x where f(x) changes sign. There are actually two reasons that f(x) could change sign: a root, or an asymptote. In your case there are three roots, and three asymptotes. Success here depends on having a small enough increment in x so that you don't completely miss a crossing. Based in the graph above it looks like 0.001 is small enough.
Here, y is a vector which contains the sign of f (as +1 or -1) at x between -75 and 0, in increments of 0.001. The limits (-75,0) were chosen by inspecting the plot above. We can see visually that there are three roots. The first two cross from (+) to (-), and the third crosses from (-) to (+). So we identify the index of x where the crossings occur (using which(...)), and then create a vector that contains the first two elements of plus.to.minus and the third element of minus.to.plus. Then we call uniroot(...) using increment=c(x[i],x[i+1]) where i is the index of the appropriate crossing.
Finally, we plot the results to confirm that we have in fact found the roots. This is really important - always, always plot the results. It turns out that uniroot(...) will find a "root" where there is an asymptote, so you have to make sure you've found actual roots.

Answer 2

Use uniroot() to solve equations in one variable:
f <- function(x){
1/8 * (1/(12+x)) + 1/4*(1/(40+x)) + 3/4*(1/(50+x)) - 1/20
}
uniroot(f, interval = c(-1e+08, 1e+08))
Notice that in the function, f, I subtract 1/20. This is because uniroot() finds the zero of the function.
In this case, you will get the error:
Error in uniroot(f, interval = c(-1e+08, 1e+08)) :
f() values at end points not of opposite sign
To correct this, you need to make sure the zero exists and if it does, move the interval, (a, b) so that f(a) == -f(b)