0
$\begingroup$

It's been a while since I studied maths, so hopefully i am using the right terms. I want to solve the following equation for $t$. That is, I want an equation like $t =$ .....

I tried factoring it and whatnot and wound up with it in quadratic equation form. But solving the quadratic equation gave values of $t$ that are impossible for any given $v$. I'm certain I solved the quadratic equation properly, so i can only assume I got to the quadratic equation in an erroneous way. Any help would be greatly appreciated. Cheers.

$\left(\frac{330}{t}\right)^2 = \sqrt{1.36 + \left(1.17 + \frac{v}{t} \right)^2}$

  • 0
    It is not possible to find an explicit formula for a quartic even cubic equation always.2017-02-28
  • 0
    I don't know what that means.2017-02-28

1 Answers 1

1

Let $x=1/t$; then the equation can be rewritten as

$$(330x)^2=\sqrt{1.36+(1.17+vx)^2}.$$

Squaring both sides yields

$$(330x)^4=1.36+(1.17+vx)^2,$$

and expanding and rearranging gives us

$$11859210000\cdot x^4-v^2\cdot x^2-2.34v\cdot x-2.7289=0.$$

This is a quartic, not a quadratic equation. There are closed formulas for the solutions, but I suppose numeric approaches are what you're looking for?

  • 0
    I just want to be able to solve t for a given v. However that happens I'm not too fussed.. ;)2017-02-28
  • 0
    It suffices to solve this quartic, and check which solutions satisfy the initial equation.2017-02-28
  • 0
    How do I isolate x though?2017-02-28
  • 0
    There are very complicated and ugly formulas for the exact solutions of quartic equations. See [this link](https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots). If what you want is numeric solutions, then pick your favorite method and go for it.2017-02-28
  • 0
    Ok, that looks like Chinese to me. Am I right in understanding that there are multiple solutions? In that case, any negative answers can be excluded as they are impossible in the case I am looking into (t= time). Can someone solve for the other cases?2017-02-28
  • 0
    Yes, a polynomial of degree $n$ may have up to $n$ distinct solutions.2017-02-28
  • 0
    So can you or anyone else compute some (all?) of those solutions?2017-02-28