Mathematics - Quadratic Equation Question with Solution | TestHub

Let $y=\frac{x^2-x+1}{x^2+x+1}$. Cross-multiplying gives $y(x^2+x+1)=x^2-x+1$, which rearranges to $(y-1)x^2+(y+1)x+(y-1)=0$. For $x$ to be real, the discriminant $D$ of this quadratic in $x$ must be non-negative. If $y=1$, the equation becomes $2x=0$, so $x=0$, which is a real solution. Thus $y=1$ is in the range. If $y\ne 1$, then $D=(y+1{)}^2-4(y-1)(y-1)\ge 0$. This simplifies to $(y+1{)}^2-4(y-1{)}^2\ge 0$. Using the difference of squares formula, $((y+1)-2(y-1))((y+1)+2(y-1))\ge 0$. This becomes $(-y+3)(3y-1)\ge 0$, or $(y-3)(3y-1)\le 0$. The roots are $y=1/3$ and $y=3$. So, $1/3\le y\le 3$. Since $y=1$ is included in this interval, the range of $y$ is $[1/3,3]$.