Solve for the roots for the equation `X^4-X^3+X^2-X+1=0.`

Expert Answers

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This equation has no real roots. Let's prove this.

 

Split the term `x^2` into halves and divide them between the leading and trailing terms:

`x^4-x^3+x^2-x+1=(x^4-x^3+1/2 x^2)+(1/2 x^2-x+1)=`

`=x^2(x^2-x+1/2)+(1/2 x^2-x+1).`

 

Both quadratic trinomials in parentheses are always positive because their discriminants are equal to `1-4*1/2=-1` (negative) and the leading coefficients are positive. And the factor `x^2` is non-negative. So the sum is always positive and never becomes zero.

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