Problem #368
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368
Find the set of values of \(x\) for which \(x^2 + 6x + 2 > 9\).
Solution
Start with the inequality:
\(x^2 + 6x + 2 > 9\)
Simplify it:
\(x^2 + 6x + 2 - 9 > 0\)
\(x^2 + 6x - 7 > 0\)
Factor the quadratic expression:
\((x + 7)(x - 1) > 0\)
The critical points are \(x = -7\) and \(x = 1\).
Test intervals around the critical points:
1. For \(x < -7\), choose \(x = -8\):
\((-8 + 7)(-8 - 1) > 0\)
\((-1)(-9) > 0\)
\(9 > 0\) (True)
2. For \(-7 < x < 1\), choose \(x = 0\):
\((0 + 7)(0 - 1) > 0\)
\(7(-1) > 0\)
\(-7 > 0\) (False)
3. For \(x > 1\), choose \(x = 2\):
\((2 + 7)(2 - 1) > 0\)
\(9(1) > 0\)
\(9 > 0\) (True)
Thus, the solution is \(x > 1\) or \(x < -7\).