Mathematics - Quadratic Equation Question with Solution | TestHub

Let $y=\frac{x^2+2x+3}{x^2+x+1}$. We want to find the range of $y$ for real $x$. Rearranging the equation to form a quadratic in $x$: $y(x^2+x+1)=x^2+2x+3$.$y{x}^2+yx+y=x^2+2x+3⟹(y-1)x^2+(y-2)x+(y-3)=0$.

For real values of $x$, the discriminant $D$ of this quadratic equation must be non-negative, i.e., $D\ge 0$.$D=(y-2{)}^2-4(y-1)(y-3)\ge 0$.$(y^2-4y+4)-4(y^2-4y+3)\ge 0$.$y^2-4y+4-4y^2+16y-12\ge 0⟹-3y^2+12y-8\ge 0$. Multiplying by $-1$ and reversing the inequality sign: $3y^2-12y+8\le 0$.

To find the values of $y$ that satisfy this inequality, we find the roots of the quadratic equation $3y^2-12y+8=0$. Using the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $y=\frac{12\pm \sqrt{(-12{)}^2-4(3)(8)}}{2(3)}=\frac{12\pm \sqrt{144-96}}{6}=\frac{12\pm \sqrt{48}}{6}$. Since $\sqrt{48}=4\sqrt{3}$, we have $y=\frac{12\pm 4\sqrt{3}}{6}=\frac{6\pm 2\sqrt{3}}{3}$.

Since the parabola $3y^2-12y+8$ opens upwards (coefficient of $y^2$ is positive), the inequality $3y^2-12y+8\le 0$ holds when $y$ is between or equal to its roots. Therefore, the range of $y$ is $\left[\frac{6-2\sqrt{3}}{3},\frac{6+2\sqrt{3}}{3}\right]$.