If \(x\) is an integer, is \(|x| \gt 1\)?
Is \(|x| \gt 1\)?
Is \(x < -1\) or \(x > 1\)?
So, for an YES answer, \(x\) can be any integer except -1, 0, and 1.
(1) \((1 - 2x)(1 + x) \lt 0\).
Rewrite as \((2x - 1)(x + 1) > 0\) (so that the coefficient of \(x^2\) is positive after expanding): the roots are \(x=-1\) and \(x=\frac{1}{2}\). The "\(>\)" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root; therefore, the given inequality holds true for: \(x < -1\) and \(x > \frac{1}{2}\). However, \(x\) could still equal 1, so this statement is not sufficient.
(2) \((1 - x)(1 + 2x) \lt 0\).
Rewrite as \((x - 1)(2x + 1) > 0\): the roots are \(x=-\frac{1}{2}\) and \(x=1\). The "\(>\)" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root; therefore, the given inequality holds true for: \(x < - \frac{1}{2}\) and \(x > 1\). However, \(x\) could still equal -1, so this statement is not sufficient.
(1)+(2) The intersection of the ranges from (1) and (2) is \(x < -1\) and \(x > 1\). This is sufficient to determine that \(|x| > 1\).
P.S. For a more detailed explanation of the approach used to solve this question, please refer to the following link:
Solving Quadratic Inequalities: Graphical Approach Answer: C
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