Absolute Value (x^2-1)<8, I keep coming up with +/- isqrt7< x < +/- 3 but its wrong.

(x^2-1)<8

(x+3)(x-3)<8
-3<x<3

There should be no i (sqrt(-1)) in your answer. "greater than" and "less than" have no meaning in complex number space, unless you are talking about real magnitudes.

Your statement " x < +/- 3 " would imply, more simply, x < -3, since whatever is less than -3 is also less than +3.

When x^2-1 = 8, x = + or - 3. Between those numbers, x^2 -1 is less than 8.
-3 < x < 3 is the answer.

Isn't the problem set up as -8<x^2-1<8 so -7<x^2<9 so +/-sqrt-7<x<+/-sqrt9 because it is an absolute problem?

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To solve the inequality |x^2-1| < 8, we need to consider two cases:

Case 1: x^2-1 is positive
In this case, the inequality |x^2-1| < 8 becomes (x^2-1) < 8.
First, let's solve this quadratic inequality:

x^2 - 1 < 8
x^2 < 9
-3 < x < 3

So for this case, the solution is -3 < x < 3.

Case 2: x^2-1 is negative
In this case, the inequality |x^2-1| < 8 becomes -(x^2-1) < 8.
Let's solve this quadratic inequality:

-(x^2-1) < 8
-x^2 + 1 < 8
-x^2 < 7
x^2 > -7

Since x^2 is always non-negative, there is no restriction on x^2. Therefore, this inequality holds true for all real values of x.

Combining the solutions from both cases, we have: x ∈ (-∞, ∞).

So the correct solution to the inequality |x^2-1| < 8 is x ∈ (-∞, ∞).