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I want to solve cubic equations with some pattern or formula. My maths teacher told me that if we want to solve cubic equations we have to find any one value for the equation which satisfy it and then divide the main equation by it. But I find that method little useless . Is there any other simplier or good method to solve it. You can explain it by an example.

Michael Rozenberg
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  • Cardano's method. – hamam_Abdallah Oct 01 '17 at 14:19
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    Disclaimer. I am not responsible of possible headaches. https://en.wikipedia.org/wiki/Cubic_function#General_solution_to_the_cubic_equation_with_real_coefficients – Raffaele Oct 01 '17 at 14:19
  • It is some what difficult is there any easy one –  Oct 01 '17 at 14:22
  • The cubic equations that come up in beginning algebra classes are usually specially arranged so that your math teacher's method will work. For general cubics there is a formula (like the quadratic formula but harder). @Raffaele 's comment points to its wikipedia page. – Ethan Bolker Oct 01 '17 at 14:23
  • https://math.stackexchange.com/questions/843518/solving-a-cubic-polynomial-equation https://math.stackexchange.com/questions/2377324/how-to-find-roots-of-cubic-equation https://math.stackexchange.com/questions/61725/is-there-a-systematic-way-of-solving-cubic-equations – MCCCS Oct 01 '17 at 14:24
  • Is it very difficult or easy like shridharyacharya formula –  Oct 01 '17 at 14:25
  • If you have integer coefficients and the leading coefficient is $1$ (usually the case in such exercises), you only need to check the positive and negative divisors of the constant term (If the constant term is $0$, you can factor $x$ out) to find possible integer roots. Otherwise, if you want to avoid the complicated formula, use numerical methods to find the roots. – Peter Oct 01 '17 at 14:29
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    Its really a headache. –  Oct 01 '17 at 14:30
  • @AtharvaShetty Even worse, in the case all roots are real the Cardano-method requires complex radicals! You can avoid this using geometric methods. My suggestion however is to use numerical methods in the case you cannot guess a root. If you can guess one, apply polynomial division. – Peter Oct 01 '17 at 14:34
  • @AtharvaShetty By the way, for degree $4$ there is a formula as well, but it is really awful, if you do not use a computer. Incredibly, it has been shown that for degree $5$ and higher, no such general formula exists. – Peter Oct 01 '17 at 14:42

1 Answers1

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For example.

Solve the following equation. $$x^3-3x+1=0.$$ Solution.

Let $x=2\cos\alpha$.

Thus, $$8\cos^3\alpha-6\cos\alpha+1=0$$ or $$\cos3\alpha=-\frac{1}{2}$$ or $$3\alpha=\pm120^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$ or $$\alpha=\pm40^{\circ}+120^{\circ}k,$$ which gives the answer: $$\{2\cos40^{\circ},2\cos80^{\circ},-2\cos20^{\circ}\}.$$

Cubic equations are not so easy thing.

For example, one of roots of the equation $$x^3+x^2-10x-8=0$$ it's $$x_1=2\left(\cos\frac{2\pi}{31}+\cos\frac{4\pi}{31}+\cos\frac{8\pi}{31}+\cos\frac{16\pi}{31}+\cos\frac{30\pi}{31}\right).$$

Michael Rozenberg
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  • But in this eqn we want real roots not in cos –  Oct 01 '17 at 14:34
  • @Atharva Shetty $\cos20^{\circ}$ is a real number, but we can not write it in radicals. For example $\sqrt2$ we can write in radicals, but $\ln2$ or $\cos20^{\circ}$ we can not, but $\ln2$ is a real number. – Michael Rozenberg Oct 01 '17 at 14:38