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Real solution of \(x^3 +1=x \)

I was just wondering about its different solutions while one which I did was Cardan's solution

\(x^3 \)-\(x \)+1=0

Let \(x \)=\(y + z \) and \(yz \)=\(\frac {1}{3} \) \(\Rightarrow \) \(3yz - 1 \)=0

\(x^3 \)=\((y+z)^3 \) =\(y^3 + z^3 + 3yz (y+z) \)=\(y^3 + z^3 +3yzx \)

Putting in the equation value of \(x^3 \)

\(y^3 + z^3 + (3yz-1)x +1 \)

\(y^3 + z^3 +1 \) \(\Rightarrow \) \(y^3+z^3 \)=-1 and also \(yz \)=\(\frac {1}{3} \) \(\Rightarrow \) \(y^3z^3 \)=\(\frac {1}{27} \)

Making quadratic in \(y^3 \) and \(z^3 \) solving we can get y and z and thus y+z=x by making it as factor we can get other solution which are complex.

Can you provide such simillar solution and what if question was \(x^4 \)+1=\(x \)

Note by Tarun Garg
4 years ago

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The simplest way to solve this quartic, without wheeling in a load of theory, is to try to complete the square. We can write \[ x^4 - x + 1 \; = \; (x^2 + a)^2 - b(x + c)^2 \] provided that \(b=2a\), \(2bc = 1\) and \(a^2 - bc^2 = 1\). Then \(b = 2a\), \(c = \tfrac{1}{4a}\), and \(a\) satisfies the equation \[ a^2 - \tfrac{1}{8a} \; = \; 1 \] Thus we need to solve a cubic equation to find \(a\). Once we have done this, we can use \(a\), \(b\) and \(c\) to factorise \(x^4 - x + 1\) as a product of two quadratics, and then solving the equation \(x^4 - x + 1 = 0\) is straightforward.

It is possible to solve all quartics with no \(x^3\) term this way; completing the square aims to write the quartic as a difference of two squares, and the condition that needs to be solved to make this possible requires us to solve a cubic equation. If we can solve cubics, we can solve quartics. The restriction that the quartic should have no \(x^3\) term is no problem; any cubic can be converted into one without an \(x^3\) term simply by translation, regarding it as a function of \(x+u\) for a suitable \(u\).

Mark Hennings - 4 years ago

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A more theoretically elegant way to solve the quartic is the following. If \(t_1,t_2,t_3,t_4\) are the roots of the quartic \[ x^4 + ax^3 + bc^2 + cx + d = 0 \] then consider \[ s_1 = t_1t_2+ t_3t_4 \qquad s_2 = t_1t_3+t_2t_4 \qquad s_3 = t_1t_4+t_2t_3 \] We can use the usual polynomial root techniques to identify \(s_1+s_2+s_3 (= b)\), \(s_1s_2+s_1s_3+s_2s_3\) and \(s_1s_2s_3\) in terms of \(a,b,c,d\), and hence find a cubic equation whose roots are \(s_1,s_2,s_3\). Once we know the values of \(s_1,s_2,s_3\), finding the values of \(t_1,t_2,t_3,t_4\) is not difficult.

Mark Hennings - 4 years ago

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That's a really good method getting lots of things to think here.

Tarun Garg - 4 years ago

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Real solution

\[x={\left( \frac{\sqrt{23}}{2\,{3}^{\frac{3}{2}}}-\frac{1}{2}\right) }^{\frac{1}{3}}+\frac{1}{3\,{\left( \frac{\sqrt{23}}{2\,{3}^{\frac{3}{2}}}-\frac{1}{2}\right) }^{\frac{1}{3}}}\]

Joseph Gomes - 4 years ago

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any process for solution of x^5+1=x

Ankan Ghosh - 4 years ago

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No. The quintic polynomial equation \(x^5 - x +1 = 0\) cannot be solved by these techniques (adding, subtracting, multiplying, dividing, taking \(n\)th roots). Collectively, these techniques are called "using radicals". While we can solve quadratics, cubics and quartics by radicals, we cannot (in general) solve quintics by radicals. This result is one of the great triumphs of Galois Theory, and relies on the fact that the permutation group \(S_5\) of five symbols is a much more complicated group (in a particular sense) than the permutation group \(S_4\) of four symbols.

Mark Hennings - 4 years ago

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