To solve the quadratic inequality \( x^2 + x - 42 \geq 0 \) by factoring, we first need to factor the quadratic expression.
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Factoring the quadratic: We need to find two numbers that multiply to \(-42\) (the constant term) and add up to \(1\) (the coefficient of \(x\)). The numbers \(7\) and \(-6\) work because:
\[ 7 \cdot (-6) = -42 \quad \text{and} \quad 7 + (-6) = 1 \]
Therefore, we can factor the quadratic as:
\[ (x - 6)(x + 7) \geq 0 \]
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Finding critical points: Set the factored expression to zero to find the critical points:
\[ (x - 6) = 0 \quad \Rightarrow \quad x = 6 \] \[ (x + 7) = 0 \quad \Rightarrow \quad x = -7 \]
Thus, the critical points are \(x = -7\) and \(x = 6\).
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Testing intervals: The critical points divide the real number line into three intervals:
- \( (-\infty, -7) \)
- \( (-7, 6) \)
- \( (6, \infty) \)
We will determine the sign of the product \((x - 6)(x + 7)\) in each interval.
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For \( x \in (-\infty, -7) \) (e.g., \(x = -8\)): \[ (-8 - 6)(-8 + 7) = (-14)(-1) = 14 \quad (> 0) \]
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For \( x \in (-7, 6) \) (e.g., \(x = 0\)): \[ (0 - 6)(0 + 7) = (-6)(7) = -42 \quad (< 0) \]
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For \( x \in (6, \infty) \) (e.g., \(x = 7\)): \[ (7 - 6)(7 + 7) = (1)(14) = 14 \quad (> 0) \]
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Combining results: We found that \((x - 6)(x + 7) \geq 0\) in the intervals:
- \( (-\infty, -7) \)
- \( (6, \infty) \)
Additionally, at the critical points \(x = -7\) and \(x = 6\), the product is zero (i.e., \((x - 6)(x + 7) = 0\)), so we include these points.
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Final solution: Therefore, the solution to the inequality \(x^2 + x - 42 \geq 0\) in interval notation is:
\[ (-\infty, -7] \cup [6, \infty) \]