Question 362883
By the rational roots theorem, if the equation has rational roots, they are of the form,
{{{x=1/1}}} or {{{x=-1/1}}}
For {{{x=1}}},
{{{1^2+1-1=1+0=1}}} so that is not a solution.
For {{{x=-1}}},
{{{1^2-1-1=1-2=-1}}} so that is not a solution.
So there are no rational roots.
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{{{graph(300,300,-3,3,-3,3,x^2+x-1)}}}