Question

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

Answer 1

(x^2-1)<8
(x+3)(x-3)<8
-3<x<3

Answer 2

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.

Answer 3

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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