Solve x^6+1=x^4+x^2

$${ x }^{ 6 } +1= { x }^{ 4 } + { x }^{ 2 }$$

Solve for x

$$x=-1$$

$$x=1$$

Solution Steps

Subtract x^{4} from both sides.

$$x^{6}+1-x^{4}=x^{2}$$

Subtract x^{2} from both sides.

$$x^{6}+1-x^{4}-x^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{6}-x^{4}-x^{2}+1=0$$

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^{6}-x^{4}-x^{2}+1 by x-1 to get x^{5}+x^{4}-x-1. Solve the equation where the result equals to 0.

$$x^{5}+x^{4}-x-1=0$$

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^{5}+x^{4}-x-1 by x-1 to get x^{4}+2x^{3}+2x^{2}+2x+1. Solve the equation where the result equals to 0.

$$x^{4}+2x^{3}+2x^{2}+2x+1=0$$

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}+2x^{3}+2x^{2}+2x+1 by x+1 to get x^{3}+x^{2}+x+1. Solve the equation where the result equals to 0.

$$x^{3}+x^{2}+x+1=0$$

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^{3}+x^{2}+x+1 by x+1 to get x^{2}+1. Solve the equation where the result equals to 0.

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

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, 0 for b, and 1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-4}}{2}$$

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.

$$x=1 x=-1$$

Solve for x (complex solution)

$$x=i$$

$$x=-i$$

$$x=-1$$

$$x=1$$