Q.Find the range of the expression $\dfrac{x^2 + x + 1}{x^2 - x + 1}$.

Set $y$ equal to the expression, cross-multiply into a quadratic in $x$, and demand a real discriminant — that inequality in $y$ gives the range.

Let

$y = \frac{x^2+x+1}{x^2-x+1}$

The denominator $x^2-x+1$ has discriminant $1-4=-3<0$, so it's never zero — the function is defined for all real $x$.

Cross-multiplying:

$y(x^2-x+1) = x^2+x+1$

$(y-1)x^2 - (y+1)x + (y-1) = 0$

Case $y=1$: the equation becomes $-2x=0\implies x=0$, which is a valid real solution (check: at $x=0$, the expression is $\tfrac{1}{1}=1$). So $y=1$ is attained.

Case $y\ne1$: this is a genuine quadratic in $x$, and for $x$ to be real we need discriminant $\ge0$:

$(y+1)^2 - 4(y-1)^2 \ge 0$

Expand:

$(y^2+2y+1) - 4(y^2-2y+1) \ge 0 \implies -3y^2+10y-3\ge0 \implies 3y^2-10y+3\le0$

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