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1.3 When we solve the following quadratic inequality: \( x^{2}-1>0 \) and approach it like a linear inequality, the following happens as a solution: \[ x^{2}-1>0 \] \[ (x+1)(x-1)>0 \] \[ x+1>0 \text { or } x-1>0 \] \[ x>-1 \text { or } x>1 \] \( \begin{array}{l}\text { 1.3.1 } \begin{array}{l}\text { Represent this solution above on a number line. } \\ \text { 1.3.2 } \begin{array}{l}\text { Does the solution on the number line make sense, explain your } \\ \text { answer. }\end{array} \\ \text { (2) }\end{array}\end{array} \)

Ask by Hamilton Vargas. in South Africa
Jan 30,2025

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Tutor-Verified Answer

Answer

The solution to \( x^{2} - 1 > 0 \) is \( x < -1 \) or \( x > 1 \). On the number line, this means all numbers less than \(-1\) and all numbers greater than \(1\) are shaded.

Solution

Sure, let's tackle each part of the problem step by step. --- ### **1.3.1 Representing the Solution on a Number Line** Given the inequality: \[ x^{2} - 1 > 0 \] The user approached it by treating it as a linear inequality: \[ (x + 1)(x - 1) > 0 \\ x + 1 > 0 \quad \text{or} \quad x - 1 > 0 \\ x > -1 \quad \text{or} \quad x > 1 \] **Solution Representation:** Based on the user's approach, the solution is \( x > -1 \) **or** \( x > 1 \). However, since \( x > 1 \) is already included in \( x > -1 \), the combined solution simplifies to: \[ x > -1 \] **Number Line Representation:** ``` <---------------------|====================> -1 → ``` - **Shaded Region:** All numbers to the right of \(-1\) (excluding \(-1\) itself if the inequality is strict). - **Point \(-1\):** Typically represented with an open circle to indicate that \(-1\) is not included in the solution. --- ### **1.3.2 Evaluating the Solution on the Number Line** **Does the Solution Make Sense?** No, the solution obtained by treating the quadratic inequality as a linear one does **not** fully make sense. Here's why: 1. **Incorrect Approach:** The user applied the rules for **linear inequalities** to a **quadratic inequality**, which requires a different approach. 2. **Actual Solutions for Quadratic Inequality:** - The inequality \( x^{2} - 1 > 0 \) actually factors to \( (x + 1)(x - 1) > 0 \). - For the product to be positive, both factors must be **positive** or both must be **negative**. - **Both Positive:** \( x + 1 > 0 \) **and** \( x - 1 > 0 \) ⇒ \( x > 1 \). - **Both Negative:** \( x + 1 < 0 \) **and** \( x - 1 < 0 \) ⇒ \( x < -1 \). - **Correct Solution:** \( x < -1 \) **or** \( x > 1 \). 3. **Discrepancy:** - The user's approach only captures \( x > -1 \), missing the \( x < -1 \) part of the solution. - **Correct Number Line Representation Should Be:** ``` <====|---------------------|====================> -1 1 → ``` - **Shaded Regions:** - All numbers to the left of \(-1\) ( \( x < -1 \) ). - All numbers to the right of \(1\) ( \( x > 1 \) ). - **Points \(-1\) and \(1\):** Both typically represented with open circles. **Conclusion:** The solution obtained by treating the quadratic inequality as a linear one is incomplete and **incorrect**. The proper method for solving quadratic inequalities involves analyzing the intervals determined by the roots of the equation and testing each interval to determine where the inequality holds true. This ensures that all valid solutions are captured.

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Bonus Knowledge

To represent the solution \( x^2 - 1 > 0 \) on a number line, we first find the critical points where \( x^2 - 1 = 0 \), which are \( x = -1 \) and \( x = 1 \). These points divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \). Now, we check the signs of the expression \( (x + 1)(x - 1) \) in each interval. - For \( x < -1 \) (choose, say, \( x = -2 \)): \( (-2 + 1)(-2 - 1) > 0 \) which is true. - For \( -1 < x < 1 \) (choose, say, \( x = 0 \)): \( (0 + 1)(0 - 1) < 0 \) which is false. - For \( x > 1 \) (choose, say, \( x = 2 \)): \( (2 + 1)(2 - 1) > 0 \) which is true. Thus, the solution to the inequality \( x^2 - 1 > 0 \) is represented on the number line with open circles at -1 and 1, and shaded areas extending to both \( (-\infty, -1) \) and \( (1, \infty) \). The solution on the number line makes sense because we are looking for the intervals where \( (x + 1)(x - 1) \) is positive. In the intervals \( (-\infty, -1) \) and \( (1, \infty) \), the product is positive, indicating that \( x \) must be either less than -1 or greater than 1 for the original inequality \( x^2 - 1 > 0 \) to hold true. The interval \((-1, 1)\) does not satisfy the inequality, confirming that our solution is accurate!

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