Solution for the given rational inequality x2-1x+2 \geq 0.

Given: ${\displaystyle x2-1x+2\ge 0}$.
Solving rational inequality means finding the values of x which will make the inequality TRUE. We need to set numerator and denominator to zero and solve for x to find the boundary points. Based on the boundary points, identify the possible solution intervals of the inequality, then pick test point from each possible solution intervals and check whether it makes the original inequality true or false. The interval which contains the test points making the inequality TRUE are the solution set of the given inequality.
Calculation:
$\begin{array}{|cc|}\hline Description& \text{Solving Equation}x2-1x+2\ge 0\\ Step1:\text{Whenever we have and expression of the form ax+bcx+d, We need to set numerator and denominator to zero and solve for x to find the boundary points.}& x2-1=0(or)x+2=0(x-1)(x+1)=0(or)x+2=0x=-1(or)x=1(or)x=-2\\ Step2:\text{The boundary points of this inequality are found by setting up left side factored expression to zero and these point -2 are not included because it will make the denominator zero and the point 1 and -1 are included because of}\ge \text{sign. List the possible solution set.}& \text{possible solution set}(-\mathrm{\infty },-2),(-2,-1),[-1,1],[1,\mathrm{\infty })\\ Step3:\text{Picking -5 as test point in interval}(-\mathrm{\infty },-2)\text{and checking whether it makes the inequality true}& (-5)2-1(-5)+2\ge 0FALSE\\ Step4:\text{Picking -32 as test point in interval (-2,-1) and checking whether it makes the inequality true}& (-32)2-1(-32)+2\ge 0\Rightarrow (54)(12)\ge 0TRUE\\ Step5:\text{Picking 0 as test point in interval (-1,-1) and checking whether it makes the inequality true}& (0)2-1(0)+2\ge 0\Rightarrow (-1)(2)\ge 0FALSE\\ Step6:\text{Picking 2 as test point in interval}(-1,\mathrm{\infty })\text{and checking whether it makes the inequality true}& 22-1(2)+2\ge 0\Rightarrow (3)(4)\ge 0TRUE\\ \hline\end{array}$
Conclusion:
The solution for the given rational inequality is ${\displaystyle (-\mathrm{\infty },-2)}$ or ${\displaystyle [1,\mathrm{\infty })}$.