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FORENOTE: My question relates to finding the eigenvalues of a non linear map evaulauted at a fixed point in order to solve for the bifurcations of the dynamical system wrt a parameter c, although I am interested purely in the mathematics of my question.

The expression I am trying to solve (for c) is as follows:
$\frac{9 + 4c \pm{\sqrt{225 + 48c + 16c^2}}}{12} = 1$
This can be simplified to a root finding problem
$-3 + 4c \pm{\sqrt{225 + 48c + 16c^2}} = 0$

To my understanding there should be 2 or maybe even 4 answers (complex answers included) for c however I am only able to find 1. The answer I am able to find is somewhat trivial and comes from rearranging the above equation and squaring both sides to find c = -3:
$225 + 48c + 16c^2 = (3-4c)^2 \therefore c = -3$

At the very least it seems that there should be two answers for c, one for the $+$ case of $\pm$ and the other for the $-$ case, but I do not know how to find them. Indeed there should be two answers because c = -3 is only valid for the $+$ case (upon evaluating the 2nd expression for the $-$ case the answer is $-30$ which is not $0$ i.e. roots are not repeated). Intuition also tells me that, since there is a quadratic involved in the square root, each of these cases may even have two solutions for c.
In essence I am looking for a mathematical way of finding the other answer(s) for c, or an explanation as to why it does not exist.

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  • $\begingroup$ Thats the game isn't it... it wouldn't be mathematics if we were only interested in real solutions :) $\endgroup$ Commented Jul 22, 2022 at 13:33
  • $\begingroup$ See WolframsAlpha's take on this. $\endgroup$ Commented Jul 22, 2022 at 16:17
  • $\begingroup$ @MatthewEdizBeadman "there should be two answers for $c$" $\;-\;$ No, there is only one root, and you just proved that. Squaring an equation can introduce extraneous roots, but it never discards legitimate roots of the original equation. $\endgroup$
    – dxiv
    Commented Jul 22, 2022 at 20:50
  • 1
    $\begingroup$ It may be clearer to work this backwards: start with $m=0$. Add some quantity $x^2$, so $x^2+m = x^2$. This has an infinity of solutions for $x$. Now take a square root: $\sqrt{x^2+m} = x$. This has the same number of solutions. $x^2+m=x^2$ is not a quadratic equation, and the only conclusion is that $m=0$. $\endgroup$ Commented Jul 23, 2022 at 11:53
  • $\begingroup$ Thank you all for your help, I guess I couldn't convince myself that there could be no answer for the - case ($-c+4c-\sqrt{225+48c+16c^2}=0$), but im happy to move on now :) $\endgroup$ Commented Jul 23, 2022 at 14:56

1 Answer 1

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$$225+48x+16 x^2 - (3-4x)^2=0$$ has no term in $x^2$, the two terms cancel each other out, leaving you with a linear equation in $x$.

A numeric search for a complex solution for the negative case came up with nothing:

package main

import (
    "fmt"
    "math"
    "math/cmplx"
)

func main() {
    limit := 1000.0
    step := 0.1
    min := math.Inf(1)
    for x := -limit; x <= limit; x += step {
        for y := -limit; y <= limit; y += step {
            c := complex(x, y)
            ex := complex(3, 0) + 4*c - cmplx.Sqrt(complex(225, 0)+48*c+16*c*c)
            a := cmplx.Abs(ex)
            if a < min {
                if a < 100 {
                    fmt.Printf("%g %v %v\n", a, c, ex)
                }
                min = a
            }
        }
    }
}

The minimum this achieved on my computer is this:

3.0000930225447293 (-1.4999999998411289-1000i) (-3.000000000000004-0.023625069768058893i)

This is Google Go language, you can run it in the browser here: https://go.dev/play/p/yEYdLxdm9on However, it will time out, so you will need to change the limits.

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  • $\begingroup$ indeed, but if you try to input that solution into the minus case it does not give you the correct answer of 1 (relating to the first equation) and therefore the solution is incomplete. I know that the second solution will be complex, I just dont know how to find it. In a sense information is lost when you square both sides, and I am curious how to regain that information. $\endgroup$ Commented Jul 22, 2022 at 13:30
  • $\begingroup$ @MatthewEdizBeadman a numeric search for a complex solution from -1000 to 1000 comes up with nothing whatsoever, and it seems to reach a minimum of $-3+0i$ as the imaginary part goes to infinity. I've added the code here so you can check yourself. $\endgroup$ Commented Jul 22, 2022 at 22:19

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