Smallest positive root of $x^3 - 5x + 1 = 0$ (Forbidden to use Cardanos method)?

Question

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!

Answer 1

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.

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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, $(0,\sqrt{5/3})$.