Problem 66 - Algebra and Trigonometry

Question

Solve each inequality in Exercises \(65-70\) and graph the solution set on a real number line. $$ \left|x^{2}+6 x+1\right|>8 $$

Step-by-Step Solution

Answer

The solution to the inequality \( \left|x^{2}+6 x+1\right| > 8 \) is (-∞, -7) U (1, ∞). This is represented on a number line with open circles at x = -7 and x = 1, marking the ranges of x-values that satisfy the inequality.

Step 1: Set Up Two Inequalities

Because the absolute value of a number is either positive or negative, set two inequalities: \(x^{2}+6x+1 - 8 > 0\) and \(-x^{2}-6x-1 + 8 < 0 \)

Step 2: Simplify the Inequalities

Simplify inequalities from step 1 to \( x^{2}+6x-7 > 0\) and \(-x^{2}-6x+7 < 0 \)

Step 3: Factor and Solve the Inequalities

Factorize the inequalities to their simplest forms and find their roots. The factored form of \(x^{2}+6x-7\) is \((x + 7)(x - 1) > 0\). So, the roots are x=1 and x=-7. Do the same for second inequality: \(-(x - 7)(x + 1) < 0\), with roots x=-1 and x=7

Step 4: Determine the Solution Sets

Find the solution for each inequality. The solutions for \( x^{2}+6x-7 > 0 \) are two intervals: (-∞, -7) and (1, ∞). The solutions for \(-x^{2} - 6x + 7 < 0\) are (-1, 7)

Step 5: Obtain the Solution of the Original Inequality

The solution of the absolute value inequality will be the intersection of the two intervals from step 4. The intersection of (-∞, -7), (1, ∞) and (-1, 7) is (-∞, -7) U (1, ∞)

Step 6: Graph the Solution Set on the Number Line

Draw a number line, plotting the intervals (-∞, -7) and (1, ∞). An open circle is used at x = -7 and x = 1 because these endpoints are not included in the solution set