homework #14: solving quadratic inequalities | algebra II ... - Free Printable

Problem Analysis and Solution

#### Writing About Mathematics

1. Rita's Statement:
- Rita said that when the product of three linear factors is greater than zero, all of the factors must be greater than zero or all of the factors must be less than zero.
- Agreement: Yes, Rita is correct.
- Explanation:
- The product of three numbers is positive if:
1. All three numbers are positive, or
2. Two of the numbers are negative and one is positive (since the product of two negatives is positive, and multiplying by a positive number keeps it positive).
- However, in the context of linear factors, we are typically dealing with continuous intervals on the real number line. For the product of three linear factors to be positive, the signs of the factors must either all be positive or all be negative at any given point.
- This is because the sign of each factor changes at its root (where the factor equals zero), and the product's sign changes whenever an odd number of factors change sign. Therefore, for the product to remain positive, the signs of all factors must align as either all positive or all negative.

2. Shelley's Statement:
- Shelley said that if \((x-7)(x-5) < 0\), then \((x-7)\) must be the negative factor and \((x-5)\) must be the positive factor.
- Part (a): Agreement with Shelley
- Disagreement: Shelley's statement is not always correct.
- Explanation:
- The inequality \((x-7)(x-5) < 0\) means that the product of the two factors is negative. For the product of two numbers to be negative, one must be positive and the other must be negative.
- However, it is not necessarily true that \((x-7)\) is always the negative factor and \((x-5)\) is always the positive factor. The roles can switch depending on the value of \(x\):
- If \(5 < x < 7\), then \((x-5) > 0\) and \((x-7) < 0\). Here, \((x-5)\) is positive and \((x-7)\) is negative.
- If \(x < 5\) or \(x > 7\), the inequality does not hold, so we do not consider these cases.
- Therefore, Shelley's statement is incorrect because it does not account for the possibility that \((x-5)\) could be the negative factor and \((x-7)\) could be the positive factor.

- Part (b): Determining Positive and Negative Factors
- Answer: No, it is not always possible to tell which is the positive factor and which is the negative factor.
- Justification:
- As explained above, for the product of two factors to be negative, one factor must be positive and the other must be negative. However, without additional information about the value of \(x\), we cannot definitively say which factor is positive and which is negative. The roles depend on the interval in which \(x\) lies.

#### Developing Skills

For each inequality, we will solve it step-by-step and determine the solution set for \(x\) as an integer.

3. \(x^2 + 5x + 6 < 0\)
- Factorize: \(x^2 + 5x + 6 = (x+2)(x+3)\)
- Roots: \(x = -2\) and \(x = -3\)
- Test intervals: \((-\infty, -3)\), \((-3, -2)\), \((-2, \infty)\)
- For \(x \in (-\infty, -3)\), choose \(x = -4\): \((-4+2)(-4+3) = (-2)(-1) = 2 > 0\)
- For \(x \in (-3, -2)\), choose \(x = -2.5\): \((-2.5+2)(-2.5+3) = (-0.5)(0.5) = -0.25 < 0\)
- For \(x \in (-2, \infty)\), choose \(x = 0\): \((0+2)(0+3) = (2)(3) = 6 > 0\)
- Solution: \((-3, -2)\)
- Integer solutions: \(\{ -3, -2 \}\)

4. \(x^2 + 5x - 6 > 0\)
- Factorize: \(x^2 + 5x - 6 = (x+6)(x-1)\)
- Roots: \(x = -6\) and \(x = 1\)
- Test intervals: \((-\infty, -6)\), \((-6, 1)\), \((1, \infty)\)
- For \(x \in (-\infty, -6)\), choose \(x = -7\): \((-7+6)(-7-1) = (-1)(-8) = 8 > 0\)
- For \(x \in (-6, 1)\), choose \(x = 0\): \((0+6)(0-1) = (6)(-1) = -6 < 0\)
- For \(x \in (1, \infty)\), choose \(x = 2\): \((2+6)(2-1) = (8)(1) = 8 > 0\)
- Solution: \((-\infty, -6) \cup (1, \infty)\)
- Integer solutions: \(\{ \ldots, -8, -7, -6, 2, 3, \ldots \}\)

5. \(x^2 - 3x + 2 \leq 0\)
- Factorize: \(x^2 - 3x + 2 = (x-1)(x-2)\)
- Roots: \(x = 1\) and \(x = 2\)
- Test intervals: \((-\infty, 1)\), \((1, 2)\), \((2, \infty)\)
- For \(x \in (-\infty, 1)\), choose \(x = 0\): \((0-1)(0-2) = (-1)(-2) = 2 > 0\)
- For \(x \in (1, 2)\), choose \(x = 1.5\): \((1.5-1)(1.5-2) = (0.5)(-0.5) = -0.25 < 0\)
- For \(x \in (2, \infty)\), choose \(x = 3\): \((3-1)(3-2) = (2)(1) = 2 > 0\)
- Solution: \([1, 2]\)
- Integer solutions: \(\{ 1, 2 \}\)

6. \(x^2 - 7x + 10 > 0\)
- Factorize: \(x^2 - 7x + 10 = (x-2)(x-5)\)
- Roots: \(x = 2\) and \(x = 5\)
- Test intervals: \((-\infty, 2)\), \((2, 5)\), \((5, \infty)\)
- For \(x \in (-\infty, 2)\), choose \(x = 1\): \((1-2)(1-5) = (-1)(-4) = 4 > 0\)
- For \(x \in (2, 5)\), choose \(x = 3\): \((3-2)(3-5) = (1)(-2) = -2 < 0\)
- For \(x \in (5, \infty)\), choose \(x = 6\): \((6-2)(6-5) = (4)(1) = 4 > 0\)
- Solution: \((-\infty, 2) \cup (5, \infty)\)
- Integer solutions: \(\{ \ldots, -2, -1, 0, 1, 2, 6, 7, \ldots \}\)

7. \(x^2 - x - 6 < 0\)
- Factorize: \(x^2 - x - 6 = (x-3)(x+2)\)
- Roots: \(x = 3\) and \(x = -2\)
- Test intervals: \((-\infty, -2)\), \((-2, 3)\), \((3, \infty)\)
- For \(x \in (-\infty, -2)\), choose \(x = -3\): \((-3-3)(-3+2) = (-6)(-1) = 6 > 0\)
- For \(x \in (-2, 3)\), choose \(x = 0\): \((0-3)(0+2) = (-3)(2) = -6 < 0\)
- For \(x \in (3, \infty)\), choose \(x = 4\): \((4-3)(4+2) = (1)(6) = 6 > 0\)
- Solution: \((-2, 3)\)
- Integer solutions: \(\{ -1, 0, 1, 2 \}\)

8. \(x^2 - 8x - 20 \geq 0\)
- Factorize: \(x^2 - 8x - 20 = (x-10)(x+2)\)
- Roots: \(x = 10\) and \(x = -2\)
- Test intervals: \((-\infty, -2)\), \((-2, 10)\), \((10, \infty)\)
- For \(x \in (-\infty, -2)\), choose \(x = -3\): \((-3-10)(-3+2) = (-13)(-1) = 13 > 0\)
- For \(x \in (-2, 10)\), choose \(x = 0\): \((0-10)(0+2) = (-10)(2) = -20 < 0\)
- For \(x \in (10, \infty)\), choose \(x = 11\): \((11-10)(11+2) = (1)(13) = 13 > 0\)
- Solution: \((-\infty, -2] \cup [10, \infty)\)
- Integer solutions: \(\{ \ldots, -4, -3, -2, 10, 11, \ldots \}\)

9. \(x^2 + x - 12 < 0\)
- Factorize: \(x^2 + x - 12 = (x+4)(x-3)\)
- Roots: \(x = -4\) and \(x = 3\)
- Test intervals: \((-\infty, -4)\), \((-4, 3)\), \((3, \infty)\)
- For \(x \in (-\infty, -4)\), choose \(x = -5\): \((-5+4)(-5-3) = (-1)(-8) = 8 > 0\)
- For \(x \in (-4, 3)\), choose \(x = 0\): \((0+4)(0-3) = (4)(-3) = -12 < 0\)
- For \(x \in (3, \infty)\), choose \(x = 4\): \((4+4)(4-3) = (8)(1) = 8 > 0\)
- Solution: \((-4, 3)\)
- Integer solutions: \(\{ -3, -2, -1, 0, 1, 2 \}\)

10. \(x^2 - 6x + 5 > 0\)
- Factorize: \(x^2 - 6x + 5 = (x-1)(x-5)\)
- Roots: \(x = 1\) and \(x = 5\)
- Test intervals: \((-\infty, 1)\), \((1, 5)\), \((5, \infty)\)
- For \(x \in (-\infty, 1)\), choose \(x = 0\): \((0-1)(0-5) = (-1)(-5) = 5 > 0\)
- For \(x \in (1, 5)\), choose \(x = 3\): \((3-1)(3-5) = (2)(-2) = -4 < 0\)
- For \(x \in (5, \infty)\), choose \(x = 6\): \((6-1)(6-5) = (5)(1) = 5 > 0\)
- Solution: \((-\infty, 1) \cup (5, \infty)\)
- Integer solutions: \(\{ \ldots, -2, -1, 0, 1, 6, 7, \ldots \}\)

11. \(x^2 - 2x \geq 0\)
- Factorize: \(x^2 - 2x = x(x-2)\)
- Roots: \(x = 0\) and \(x = 2\)
- Test intervals: \((-\infty, 0)\), \((0, 2)\), \((2, \infty)\)
- For \(x \in (-\infty, 0)\), choose \(x = -1\): \((-1)(-1-2) = (-1)(-3) = 3 > 0\)
- For \(x \in (0, 2)\), choose \(x = 1\): \((1)(1-2) = (1)(-1) = -1 < 0\)
- For \(x \in (2, \infty)\), choose \(x = 3\): \((3)(3-2) = (3)(1) = 3 > 0\)
- Solution: \((-\infty, 0] \cup [2, \infty)\)
- Integer solutions: \(\{ \ldots, -2, -1, 0, 2, 3, \ldots \}\)

12. \(x^2 - x < 6\)
- Rewrite: \(x^2 - x - 6 < 0\)
- Factorize: \(x^2 - x - 6 = (x-3)(x+2)\)
- Roots: \(x = 3\) and \(x = -2\)
- Test intervals: \((-\infty, -2)\), \((-2, 3)\), \((3, \infty)\)
- For \(x \in (-\infty, -2)\), choose \(x = -3\): \((-3-3)(-3+2) = (-6)(-1) = 6 > 0\)
- For \(x \in (-2, 3)\), choose \(x = 0\): \((0-3)(0+2) = (-3)(2) = -6 < 0\)
- For \(x \in (3, \infty)\), choose \(x = 4\): \((4-3)(4+2) = (1)(6) = 6 > 0\)
- Solution: \((-2, 3)\)
- Integer solutions: \(\{ -1, 0, 1, 2 \}\)

13. \(x^2 - 4x + 4 > 0\)
- Factorize: \(x^2 - 4x + 4 = (x-2)^2\)
- Roots: \(x = 2\) (double root)
- Test intervals: \((-\infty, 2)\), \((2, \infty)\)
- For \(x \in (-\infty, 2)\), choose \(x = 1\): \((1-2)^2 = (-1)^2 = 1 > 0\)
- For \(x \in (2, \infty)\), choose \(x = 3\): \((3-2)^2 = (1)^2 = 1 > 0\)
- At \(x = 2\): \((2-2)^2 = 0\)
- Solution: \((-\infty, 2) \cup (2, \infty)\)
- Integer solutions: \(\{ \ldots, -2, -1, 0, 1, 3, 4, \ldots \}\)

14. \(x^2 - 4x + 4 \geq 0\)
- Factorize: \(x^2 - 4x + 4 = (x-2)^2\)
- Roots: \(x = 2\) (double root)
- Test intervals: \((-\infty, 2)\), \((2, \infty)\)
- For \(x \in (-\infty, 2)\), choose \(x = 1\): \((1-2)^2 = (-1)^2 = 1 > 0\)
- For \(x \in (2, \infty)\), choose \(x = 3\): \((3-2)^2 = (1)^2 = 1 > 0\)
- At \(x = 2\): \((2-2)^2 = 0\)
- Solution: \((-\infty, \infty)\)
- Integer solutions: \(\{ \ldots, -2, -1, 0, 1, 2, 3, \ldots \}\)

15. \(x^2 + x - 2 < 0\)
- Factorize: \(x^2 + x - 2 = (x+2)(x-1)\)
- Roots: \(x = -2\) and \(x = 1\)
- Test intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, \infty)\)
- For \(x \in (-\infty, -2)\), choose \(x = -3\): \((-3+2)(-3-1) = (-1)(-4) = 4 > 0\)
- For \(x \in (-2, 1)\), choose \(x = 0\): \((0+2)(0-1) = (2)(-1) = -2 < 0\)
- For \(x \in (1, \infty)\), choose \(x = 2\): \((2+2)(2-1) = (4)(1) = 4 > 0\)
- Solution: \((-2, 1)\)
- Integer solutions: \(\{ -1, 0 \}\)

16. \(2x^2 - 2x - 24 \leq 0\)
- Simplify: \(x^2 - x - 12 \leq 0\)
- Factorize: \(x^2 - x - 12 = (x+3)(x-4)\)
- Roots: \(x = -3\) and \(x = 4\)
- Test intervals: \((-\infty, -3)\), \((-3, 4)\), \((4, \infty)\)
- For \(x \in (-\infty, -3)\), choose \(x = -4\): \((-4+3)(-4-4) = (-1)(-8) = 8 > 0\)
- For \(x \in (-3, 4)\), choose \(x = 0\): \((0+3)(0-4) = (3)(-4) = -12 < 0\)
- For \(x \in (4, \infty)\), choose \(x = 5\): \((5+3)(5-4) = (8)(1) = 8 > 0\)
- Solution: \([-3, 4]\)
- Integer solutions: \(\{ -3, -2, -1, 0, 1, 2, 3, 4 \}\)

17. \(2x^2 - 2x - 24 > 0\)
- Simplify: \(x^2 - x - 12 > 0\)
- Factorize: \(x^2 - x - 12 = (x+3)(x-4)\)
- Roots: \(x = -3\) and \(x = 4\)
- Test intervals: \((-\infty, -3)\), \((-3, 4)\), \((4, \infty)\)
- For \(x \in (-\infty, -3)\), choose \(x = -4\): \((-4+3)(-4-4) = (-1)(-8) = 8 > 0\)
- For \(x \in (-3, 4)\), choose \(x = 0\): \((0+3)(0-4) = (3)(-4) = -12 < 0\)
- For \(x \in (4, \infty)\), choose \(x = 5\): \((5+3)(5-4) = (8)(1) = 8 > 0\)
- Solution: \((-\infty, -3) \cup (4, \infty)\)
- Integer solutions: \(\{ \ldots, -5, -4, -3, 5, 6, \ldots \}\)

Final Answer

\[
\boxed{
\begin{array}{ll}
1. & \text{Agree with Rita.} \\
2. & \text{(a) Disagree with Shelley.} \\
& \text{(b) No, it is not always possible to tell which is positive and which is negative.} \\
3. & \{-3, -2\} \\
4. & \{\ldots, -8, -7, -6, 2, 3, \ldots\} \\
5. & \{1, 2\} \\
6. & \{\ldots, -2, -1, 0, 1, 2, 6, 7, \ldots\} \\
7. & \{-1, 0, 1, 2\} \\
8. & \{\ldots, -4, -3, -2, 10, 11, \ldots\} \\
9. & \{-3, -2, -1, 0, 1, 2\} \\
10. & \{\ldots, -2, -1, 0, 1, 6, 7, \ldots\} \\
11. & \{\ldots, -2, -1, 0, 2, 3, \ldots\} \\
12. & \{-1, 0, 1, 2\} \\
13. & \{\ldots, -2, -1, 0, 1, 3, 4, \ldots\} \\
14. & \{\ldots, -2, -1, 0, 1, 2, 3, \ldots\} \\
15. & \{-1, 0\} \\
16. & \{-3, -2, -1, 0, 1, 2, 3, 4\} \\
17. & \{\ldots, -5, -4, -3, 5, 6, \ldots\} \\
\end{array}
}
\]