Why isn't the absolute value inequality (x^2-1)</=8 set up as -8<x^2-1<8 yielding -7<x^2<9 so +/-sqrt-7<x<+/-sqrt9 so +/-isqrt7<x<+/-3?
To understand why the absolute value inequality (x^2-1) ≤ 8 is not set up as -8 < (x^2-1) < 8, let's break it down step by step:
Step 1: Start with the original absolute value inequality: (x^2-1) ≤ 8
Step 2: Recall that when dealing with absolute values, there are two cases to consider:
Case 1: When the expression inside the absolute value is positive or zero: |x| = x
If (x^2-1) is greater than or equal to zero, we can rewrite the inequality as: x^2 - 1 ≤ 8.
Step 3: Add 1 to both sides of the inequality: x^2 ≤ 9.
Step 4: Take the square root of both sides: √(x^2) ≤ √9.
Since we are dealing with the absolute value of x, we can remove the square root sign and get: |x| ≤ 3.
Case 2: When the expression inside the absolute value is negative: |x| = -x
If (x^2-1) is negative, we need to change the sign of the inequality when simplifying. In this case, we multiply both sides by -1, which gives us: -x^2 + 1 ≤ 8.
Step 5: Subtract 1 from both sides of the inequality: -x^2 ≤ 7.
Step 6: Divide both sides by -1 and reverse the inequality sign: x^2 ≥ -7.
Step 7: Since the square of any real number is non-negative, we can conclude that x is a real number without any further restrictions.
Combining the results from Case 1 and Case 2, we have the final solution: -3 ≤ x ≤ 3.
Therefore, the correct setup of the inequality (x^2-1) ≤ 8 is -3 ≤ x ≤ 3, rather than the expression you provided.