Solve the inequality: $j^{2} \geq 3 j-2$
Give your answer in interval notation. Enter DNE if there is no solution.
Answer
\(\boxed{(-\infty, 1] \cup [2, \infty)}\) is the solution to the inequality.
Step 1 ：Rearrange the inequality to the standard form of a quadratic equation: \(j^{2} - 3j + 2 \geq 0\)
Step 2 ：Find the roots of the equation using the quadratic formula, \(j = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4*1*2}}{2*1}\). The roots are 1 and 2.
Step 3 ：These roots divide the number line into three intervals: \((-\infty, 1]\), \((1, 2)\), and \([2, \infty)\).
Step 4 ：Test the sign of the inequality in each interval. Choose test values within these intervals, for example -1, 1.5, and 3.
Step 5 ：The inequality is true for the test values -1 and 3, which belong to the intervals \((-\infty, 1]\) and \([2, \infty)\) respectively.
Step 6 ：\(\boxed{(-\infty, 1] \cup [2, \infty)}\) is the solution to the inequality.
