Real solution of x3+1=xx^3 +1=x

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

x3x^3 -xx +1=0

Let xx =y+zy + z and yzyz =13\frac {1}{3} \Rightarrow 3yz13yz - 1 =0

x3x^3 =(y+z)3(y+z)^3 =y3+z3+3yz(y+z)y^3 + z^3 + 3yz (y+z) =y3+z3+3yzxy^3 + z^3 +3yzx

Putting in the equation value of x3x^3

y3+z3+(3yz1)x+1y^3 + z^3 + (3yz-1)x +1

y3+z3+1y^3 + z^3 +1 \Rightarrow y3+z3y^3+z^3 =-1 and also yzyz =13\frac {1}{3} \Rightarrow y3z3y^3z^3 =127\frac {1}{27}

Making quadratic in y3y^3 and z3z^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 x4x^4 +1=xx

Note by Tarun Garg
6 years, 10 months ago

No vote yet
8 votes

  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:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

The simplest way to solve this quartic, without wheeling in a load of theory, is to try to complete the square. We can write x4x+1  =  (x2+a)2b(x+c)2 x^4 - x + 1 \; = \; (x^2 + a)^2 - b(x + c)^2 provided that b=2ab=2a, 2bc=12bc = 1 and a2bc2=1a^2 - bc^2 = 1. Then b=2ab = 2a, c=14ac = \tfrac{1}{4a}, and aa satisfies the equation a218a  =  1 a^2 - \tfrac{1}{8a} \; = \; 1 Thus we need to solve a cubic equation to find aa. Once we have done this, we can use aa, bb and cc to factorise x4x+1x^4 - x + 1 as a product of two quadratics, and then solving the equation x4x+1=0x^4 - x + 1 = 0 is straightforward.

It is possible to solve all quartics with no x3x^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 x3x^3 term is no problem; any cubic can be converted into one without an x3x^3 term simply by translation, regarding it as a function of x+ux+u for a suitable uu.

Mark Hennings - 6 years, 10 months ago

Log in to reply

A more theoretically elegant way to solve the quartic is the following. If t1,t2,t3,t4t_1,t_2,t_3,t_4 are the roots of the quartic x4+ax3+bc2+cx+d=0 x^4 + ax^3 + bc^2 + cx + d = 0 then consider s1=t1t2+t3t4s2=t1t3+t2t4s3=t1t4+t2t3 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 s1+s2+s3(=b)s_1+s_2+s_3 (= b), s1s2+s1s3+s2s3s_1s_2+s_1s_3+s_2s_3 and s1s2s3s_1s_2s_3 in terms of a,b,c,da,b,c,d, and hence find a cubic equation whose roots are s1,s2,s3s_1,s_2,s_3. Once we know the values of s1,s2,s3s_1,s_2,s_3, finding the values of t1,t2,t3,t4t_1,t_2,t_3,t_4 is not difficult.

Mark Hennings - 6 years, 10 months ago

Log in to reply

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

Tarun Garg - 6 years, 10 months ago

Log in to reply

Real solution

x=(23233212)13+13(23233212)13x={\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 - 6 years, 10 months ago

Log in to reply

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

ankan ghosh - 6 years, 10 months ago

Log in to reply

No. The quintic polynomial equation x5x+1=0x^5 - x +1 = 0 cannot be solved by these techniques (adding, subtracting, multiplying, dividing, taking nnth 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 S5S_5 of five symbols is a much more complicated group (in a particular sense) than the permutation group S4S_4 of four symbols.

Mark Hennings - 6 years, 10 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...