Solve x^4-x^3+x=1

$$x ^ { 4 } - x ^ { 3 } + x = 1$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i+1}{2}\approx 0.5-0.866025404i$$

$$x=1$$

$$x=\frac{1+\sqrt{3}i}{2}\approx 0.5+0.866025404i$$

$$x=-1$$

Solution Steps

Subtract 1 from both sides.

$$x^{4}-x^{3}+x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-x^{3}+x-1 by x-1 to get x^{3}+1. Solve the equation where the result equals to 0.

$$x^{3}+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+1 by x+1 to get x^{2}-x+1. Solve the equation where the result equals to 0.

$$x^{2}-x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-3}}{2}$$

Solve the equation x^{2}-x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i+1}{2} x=\frac{1+\sqrt{3}i}{2}$$

List all found solutions.

$$x=1 x=-1 x=\frac{-\sqrt{3}i+1}{2} x=\frac{1+\sqrt{3}i}{2}$$

Solve for x

$$x=-1$$

$$x=1$$

Solution Steps

Subtract 1 from both sides.

$$x^{4}-x^{3}+x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-x^{3}+x-1 by x-1 to get x^{3}+1. Solve the equation where the result equals to 0.

$$x^{3}+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+1 by x+1 to get x^{2}-x+1. Solve the equation where the result equals to 0.

$$x^{2}-x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-3}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1 x=-1$$