Problem 26 - Algebra and Trigonometry

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ x^{2} \leq 2 x+2 $$

Step-by-Step Solution

Answer

The solution of the given inequality \(x^2 \leq 2x+2\) in interval notation is \((-\infty, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, \infty) \).

Step 1: Rearrange the polynomial

To solve the polynomial inequality \(x^{2} \leq 2x+2\), first, rearrange the inequality to one side which results in: \(x^2 - 2x - 2 \leq 0\)

Step 2: Factor the Polynomial

Factor the polynomial in the inequality to simplify it. In this case, the polynomial does not have factors, so we can use the quadratic formula to solve for roots. For the polynomial \(ax^2 + bx + c = 0\), the roots are found using the formula \(x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\). Applying the formula to the inequality, we'll find roots as \(x = \frac{2± \sqrt{(2)^2 - 4(1)(-2)}}{2(1)}\), which yields \(x = 1 \pm \sqrt{3}\)

Step 3: Identify Critical Numbers

The critical numbers are the roots of the polynomial and any numbers where the inequality is undefined. In this case, there are no numbers where the inequality is undefined, and the roots are \(1 - \sqrt{3}\) and \(1 + \sqrt{3}\). So, the critical numbers are \(1 - \sqrt{3}\) and \(1 + \sqrt{3}\)

Step 4: Test intervals

To find which intervals satisfy the inequality, choose a test number from each interval defined by the critical numbers and evaluate the inequality. The intervals are \((-\infty, 1 - \sqrt{3}), (1 - \sqrt{3}, 1 + \sqrt{3}), (1 + \sqrt{3}, \infty)\). An easy choice of test numbers might be -2, 1, and 2, yielding true, false, and true, respectively.

Step 5: Write final answer in interval notation

The intervals that passed the test should be joined and written in interval notation, suggest that the solution of \(x^2 - 2x - 2 \leq 0\) is \((-\infty, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, \infty) \)