Chapter 12: Problem 6
Solve the inequality: \(x^{2}-4 \leq 5\)
Short Answer
Expert verified
−3 ≤ x ≤ 3
Step by step solution
01
Rewrite the Inequality
Start by moving all terms to one side of the inequality to set it to zero. The inequality is originally given as: equation\(x^{2} - 4 \leq 5\). Move 5 to the left side to get: \(x^{2} - 4 - 5 \leq 0\).This simplifies to:\(x^{2} - 9 \leq 0\).
02
Factor the Quadratic Expression
Next, factor the quadratic expression on the left side of the inequality. This gives:\(x^{2} - 9 = (x - 3)(x + 3)\).So the inequality now is:\((x - 3)(x + 3) \leq 0\).
03
Determine the Critical Points
Identify the values of x that make the expression zero. Solve \((x - 3)(x + 3) = 0\) to get the critical points:x = 3 and x = -3.
04
Test Intervals
Determine the sign of the expression in each interval defined by the critical points. The intervals to test are: (-∞, -3), [-3, 3], and (3, ∞). 1. Test a point in (-∞, -3), e.g., x = -4:\((x - 3)(x + 3) = (-4 - 3)(-4 + 3) = (−7)(−1) = 7\) which is positive. 2. Test a point in [-3, 3], e.g., x = 0: \((x - 3)(x + 3) = (0 - 3)(0 + 3) = (−3)(3) = −9\),which is negative. 3. Test a point in (3, ∞), e.g., x = 4:\((x - 3)(x + 3) = (4 - 3)(4 + 3) = (1)(7) = 7\), which is positive. Since we want the expression to be less than or equal to zero, we select the interval [-3, 3].
05
Write the Solution
The solution to the inequality is the interval where the expression is less than or equal to zero, which we found to be: [-3, 3].Therefore,\(−3 ≤ x ≤ 3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
Factoring quadratics is a method used to break down a quadratic expression into simpler components called factors. For the expression \(x^2 - 9\), we look for two numbers that multiply to equal the constant term (-9) and add up to the coefficient of the middle term (0 in this case because there is no middle term). Since \(x^2 - 9\) fits the pattern of a difference of squares (\(a^2 - b^2 = (a - b)(a + b)\)), we can factor it as\((x - 3)(x + 3)\). This simplifies future steps by turning a single quadratic expression into a product of two binomial expressions. These factors are crucial for determining the critical points and testing intervals.
Critical Points
Critical points are the values of x that make the factored quadratic expression equal to zero. To find these points, we solve for x in the equations \((x - 3) = 0\) and \((x + 3) = 0\). Solving these equations:
- From \((x - 3) = 0\): x = 3
- From \((x + 3) = 0\): x = -3
Testing Intervals
Testing intervals involves choosing sample points from the regions divided by the critical points and evaluating the expression at those points. For our inequality, the critical points split the x-axis into three intervals:
- \((-∞, -3)\)
- \([-3, 3]\)
- \((3, ∞)\)
- In \((-∞, -3)\), we tested with x = -4: \((-4 - 3)(-4 + 3) = 7\) (positive)
- In \([-3, 3]\), we tested with x = 0: \((0 - 3)(0 + 3) = -9\) (negative)
- In \((3, ∞)\), we tested with x = 4: \((4 - 3)(4 + 3) = 7\) (positive)
Inequality Solution Steps
The final step is to bring all the results together to solve the inequality. This can be broken down into clear sub-steps per our previous explanations:
- Rewrite the inequality so that one side equals zero: \(x^2 - 9 ≤ 0\)
- Factor the quadratic expression: \((x - 3)(x + 3) ≤ 0\)
- Find the critical points by solving \((x - 3) = 0\) and \((x + 3) = 0\), resulting in x = 3 and x = -3
- Test the intervals \((-∞, -3)\), \([-3, 3]\), and \((3, ∞)\) with points from each interval
- Identify where the expression is less than or equal to zero: \([-3, 3]\)