To solve the quadratic inequality \(x^2 - 7x + 12 > 0\) by factoring, we first need to factor the quadratic expression.
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Factor the quadratic: The expression \(x^2 - 7x + 12\) can be factored as follows: \[ x^2 - 7x + 12 = (x - 3)(x - 4) \]
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Set the factors greater than zero: We need to find where \((x - 3)(x - 4) > 0\).
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Determine the critical points: The critical points occur when \( (x - 3) = 0\) and \( (x - 4) = 0\). This gives us: \[ x = 3 \quad \text{and} \quad x = 4 \]
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Test the intervals: We will test the sign of \((x - 3)(x - 4)\) in the intervals determined by the critical points \(3\) and \(4\):
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Interval 1: \( (-\infty, 3) \)
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Interval 2: \( (3, 4) \)
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Interval 3: \( (4, \infty) \)
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Test \(x = 2\) (in \( (-\infty, 3) \)): \[ (2 - 3)(2 - 4) = (-1)(-2) = 2 > 0 \]
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Test \(x = 3.5\) (in \( (3, 4) \)): \[ (3.5 - 3)(3.5 - 4) = (0.5)(-0.5) = -0.25 < 0 \]
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Test \(x = 5\) (in \( (4, \infty) \)): \[ (5 - 3)(5 - 4) = (2)(1) = 2 > 0 \]
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Conclusion from the test: We find that \((x - 3)(x - 4) > 0\) in the intervals \((-∞, 3)\) and \((4, ∞)\).
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Write the solution in interval notation: Therefore, the solution to the quadratic inequality \(x^2 - 7x + 12 > 0\) is: \[ (-\infty, 3) \cup (4, \infty) \]
Thus, in interval notation, the solutions to the quadratic inequality are \((-∞, 3) \cup (4, ∞)\).
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