Find both solutions for X in second-degree polynomial equation given Y in R

Question

I'm just starting to use R so it is possible that I'm not understanding the functions correctly.

I have length and age data from the literature and I have measured some specimens. I need to find the age of these specimens I measured based on the growth equation that I have calculated from the literature.

Here is a small example from my data:

#Example data
length <- c(0.06,0.087,0.147,0.241,0.615,1.49,2.42)
age <- c(1.3,   2.6,    3.3,    3.9,    5.45,   8,  10.5)

#Polynomial function second degree
growth <- lm(length ~ poly(age, 2, raw=TRUE)-1)

Plot of growth function:

#Need to use this equation to predict x (age) given y (length)
#New y data
mydata_length <- c(0.72,1.82,0.41,0.28)

To calculate age (x) given my data (y) I've tried multiple solutions, and I think the function spline() would be the right solution. Though, it looks like it keeps giving me only 1 of the 2 possible solutions for the polynomial equation, and it is the wrong one. Since it's a growth curve, it starts from the origin and I need the positive solutions to the equation. If I input the coefficients in another software to solve the equation, I can actually get both solutions including the one I need, that I report here. This is a very time consuming solution though, and not great for reproducibility.

#Spline doesn't give the correct results
xvals <- spline(x = growth$fitted.values, xout = mydata_length)$y

#xvals
#[1] -0.01614070  0.09184022 -0.05031075 -0.06537486

#Expected results
#5.88   9.08    4.56    3.85

Plot of growth function with expected results:

Is there a function within R to find the results I am looking for? I have other data that better fit to a linear regression, so a function to the same but with a linear model would be great to have too.

SOLUTION

I've found a way to obtain what I am looking for, probably not in the prettiest way but it works.

#Calculate age (x) based on my length data (y)

library(polynom)

#Save coefficients
coeffs <- growth$coefficients

#Create function for groth, 0 as intercept
growth_f <- polynomial(c(0, coeffs))

#List of specimen names - row names of y data frame
specimens <- c("sp1", "sp2", "sp3", "sp4")

#Row names for x (calculated age) data frame
rows <- c("discard","keep")

#New data frame with y (length) data
mydata_length_df <- data.frame(mydata_length, row.names = specimens)

#Matrix for calculated age results
x <- matrix(nrow = 2, ncol = nrow(mydata_length_df))
#Make data frame
x <- data.frame(x)
#Set row and column names
rownames(x) <- rows
colnames(x) <- specimens

#Loop to calculate both results of the polynomial for all specimens - second row is the one to keep
for(i in 1:4) {
  x[i] <- data.frame(solve(growth_f, mydata_length_df[i,]))
}        

x
#              sp1       sp2       sp3       sp4
#discard -4.996448 -8.189725 -3.671227 -2.965612
#keep     5.889859  9.083136  4.564638  3.859024

#Transpose to match rest of the data
x <- t(x)

#Data frame with both length and calculated age
new_results <- data.frame(length = mydata_length, calc_age = x[,2])

new_results
#      length  calc_age
#sp1   0.72    5.889859
#sp2   1.82    9.083136
#sp3   0.41    4.564638
#sp4   0.28    3.859024

Plot with expected and calculated results

Answer 1

If length predicts age, then the formula should be age ~ f(length). It is in the form predicted ~ predictors.

#Length predicts age
growth <- lm(age ~ poly(length, 2, raw=TRUE))

#Gridpoints
Length=seq(0, 2.5, length=100)
Age=predict(growth, list(length=grid_x))

#New predictors
lengths=c(0.72,1.82,0.41,0.28)
preds=predict(growth, list(length=lengths))
preds
       1        2        3        4 
5.642787 9.305444 4.204913 3.548846 

actual=c(5.88, 9.08, 4.56, 3.85)
plot(Age, Length, type="n")
lines(Age, Length, col="red")
points(preds, lengths, pch=23, col="purple")
points(actual, lengths, pch=16, col="blue")