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Solve for x (complex solution)
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x^{2}x^{2}+1-2xx^{2}-x\times 2+x^{2}\times 2=0
Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x^{2},x.
x^{4}+1-2xx^{2}-x\times 2+x^{2}\times 2=0
To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.
x^{4}+1-2x^{3}-x\times 2+x^{2}\times 2=0
To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.
x^{4}+1-2x^{3}-2x+x^{2}\times 2=0
Multiply -1 and 2 to get -2.
x^{4}-2x^{3}+2x^{2}-2x+1=0
Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.
±1
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}.
x=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^{3}-x^{2}+x-1=0
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.
±1
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}.
x=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^{2}+1=0
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=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}
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{-4}}{2}
Do the calculations.
x\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x=1
List all found solutions.