Laila12618 wrote:
I don’t understand… for statement 1 the inequality sign does not have the equal sign so none of z-1, z, z+1 can be zero.
Consequently, all three numbers, which are consecutive, must be negative in order for their product to be negative >>> z must be negative and A would look like the correct answer.
Anyone can explain why statement 1 is not sufficient?
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(z + 1)(z)(z – 1) < 0
Here is how to solve the above inequality easily. The "roots", in ascending order, are -1, 0, and 1, which gives us 4 ranges:
\(z < -1\);
\(-1 < z < 0\);
\(0 < z < 1\);
\(z > 1\).
Next, test an extreme value for \(z\): if \(z\) is some large enough number, say 10, then all three multiples will be positive, giving a positive result for the whole expression. So when \(z > 1\), the expression is positive. Now the trick: as in the 4th range, the expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive, and finally in the 1st, it'll be negative: \(\text{(- + - +)}\). So, the ranges when the expression is negative are: \(z < -1\) and \(0 < z < 1\).
P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).