I was just wondering about its different solutions while one which I did was Cardan's solution
x3-x+1=0
Let x=y+z and yz=31 ⇒ 3yz−1=0
x3=(y+z)3 =y3+z3+3yz(y+z)=y3+z3+3yzx
Putting in the equation value of x3
y3+z3+(3yz−1)x+1
y3+z3+1 ⇒ y3+z3=-1 and also yz=31 ⇒ y3z3=271
Making quadratic in y3 and z3 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 x4+1=x
#Algebra
#Math
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The simplest way to solve this quartic, without wheeling in a load of theory, is to try to complete the square. We can write x4−x+1=(x2+a)2−b(x+c)2 provided that b=2a, 2bc=1 and a2−bc2=1. Then b=2a, c=4a1, and a satisfies the equation a2−8a1=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 x4−x+1 as a product of two quadratics, and then solving the equation x4−x+1=0 is straightforward.
It is possible to solve all quartics with no x3 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 x3 term is no problem; any cubic can be converted into one without an x3 term simply by translation, regarding it as a function of x+u for a suitable u.
A more theoretically elegant way to solve the quartic is the following. If t1,t2,t3,t4 are the roots of the quartic x4+ax3+bc2+cx+d=0 then consider s1=t1t2+t3t4s2=t1t3+t2t4s3=t1t4+t2t3 We can use the usual polynomial root techniques to identify s1+s2+s3(=b), s1s2+s1s3+s2s3 and s1s2s3 in terms of a,b,c,d, and hence find a cubic equation whose roots are s1,s2,s3. Once we know the values of s1,s2,s3, finding the values of t1,t2,t3,t4 is not difficult.
Log in to reply
That's a really good method getting lots of things to think here.
Real solution
x=(232323−21)31+3(232323−21)311
any process for solution of x^5+1=x
Log in to reply
No. The quintic polynomial equation x5−x+1=0 cannot be solved by these techniques (adding, subtracting, multiplying, dividing, taking nth 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 S5 of five symbols is a much more complicated group (in a particular sense) than the permutation group S4 of four symbols.