-1
$\begingroup$

Guys how can i figure out the smallest positive root of $x^3 - 5x + 1 = 0$,

I am forbidden to to use Cardano's method, also i tried by giving trial values but it did not work out .

Any trick to this!

  • 0
    Where are you stuck exactly ? – 2017-02-24
  • 0
    Since i cannot use Cardanos method , i think a way out is to try for trial and error method to find a root and then the equation reduces to quadratic and hence i can solve it and thus can compare and get the smallest positive root...@YvesDaoust – 2017-02-24
  • 0
    so getting an initial root is where i am stuck... – 2017-02-24
  • 1
    As there are three real roots, you have three possibilities: 1. Insert the coefficients into Cardano's formula and use a calculator that can deal with complex numbers to verify which one is the smallest positive root. 2. Insert the coefficients into Cardano's formula and try to argue with angles etc. which choice of the roots of unity yields the smallest positive root. 3. Use the trigonometric method described in [Casus Irreducibilis](https://en.wikipedia.org/wiki/Casus_irreducibilis) to get the solution without using any complex numbers. – 2017-02-24
  • 0
    @BAYMAX: Cardano's method doesn't require an initial value. Did you mean *forbidden* instead of *restricted* ? – 2017-02-24
  • 0
    Yes,sorry, will edit. – 2017-02-24
  • 0
    Apologies to @DaivyaJadeja whom I made delete his question by mistake. – 2017-02-24
  • 0
    Wow a downvote! so curious i am now?? and a vote to close the question too .. wow!! – 2017-02-24

1 Answers 1

3

As there are three real roots, the trigonometric approach is required.

It starts from the identity

$$\cos3t=4\cos^3t-3\cos t,$$ so that the equation

$$4z^3-3z-c=0$$ is solved by $z=\cos t=\cos\left((\arccos c)/3\right)$ (and more generally $\cos(\pm\arccos c+2k\pi)/3)$).

To make this equation identical to the given one, a rescaling of the variable will do. We rewrite $x=z/u$

$$4\frac{x^3}{u^3}-3\frac xu-c=0,\\ x^3-\frac{3u^2}4x-\frac{cu^3}4=0$$ and identify

$$\frac{3u^2}4=5,\\\frac{cu^3}4=-1.$$

Then $u=\sqrt{20}/3$, $c=-3\sqrt{15}/50$. From $c$ you compute the $z$'s, then the $x$'s.

If Cardano is forbidden (there is ambiguity in the post), you can locate the roots by a study of the function.

There are to extrema given by

$$3x^2-5=0,\\x=\pm\sqrt{\frac53}.$$

As $f(0)=1>0$ and $f(\sqrt{5/3})\approx-3.3\cdots<0$ we can try $x=0$ as a starting approximation of Newton's iteration. Anyway, we will have to monitor that the root remains in the right range, $(0,\sqrt{5/3})$.

  • 0
    Thanks,i will try,It will take some time to digest this method . – 2017-02-24