Question
Solve, graph and give interval notation for the following inequality
(i) x²-5x/x-3≥0
(ii) x²+3x>10
(iii) 9-2|4x+1|>3
2. Express -3≤x≤1in the form |x-a|≤b
(i) x²-5x/x-3≥0
(ii) x²+3x>10
(iii) 9-2|4x+1|>3
2. Express -3≤x≤1in the form |x-a|≤b
Answers
Answered by
GPT-4o mini
Let's solve each of the inequalities step by step.
### (i) Solve \( \frac{x^2 - 5x}{x - 3} \geq 0 \)
1. **Factor the numerator**:
\[
x^2 - 5x = x(x - 5)
\]
So, we can rewrite the inequality as:
\[
\frac{x(x - 5)}{x - 3} \geq 0
\]
2. **Find the critical points**: Set the numerator and denominator equal to zero:
- \( x = 0 \) (from \( x \))
- \( x = 5 \) (from \( x - 5 \))
- \( x = 3 \) (from \( x - 3 \))
Thus, the critical points are \( x = 0, 3, 5 \).
3. **Test intervals**: The critical points divide the number line into the intervals: \( (-\infty, 0) \), \( (0, 3) \), \( (3, 5) \), and \( (5, \infty) \).
- Choose a test point from each interval:
- For \( x = -1 \) in \( (-\infty, 0) \):
\[
\frac{(-1)(-6)}{-4} = \frac{6}{-4} < 0
\]
- For \( x = 1 \) in \( (0, 3) \):
\[
\frac{(1)(-4)}{-2} = \frac{-4}{-2} > 0
\]
- For \( x = 4 \) in \( (3, 5) \):
\[
\frac{(4)(-1)}{1} = \frac{-4}{1} < 0
\]
- For \( x = 6 \) in \( (5, \infty) \):
\[
\frac{(6)(1)}{3} = \frac{6}{3} > 0
\]
4. **Combine the results**: The intervals where the expression is non-negative are \( [0, 3) \) and \( (5, \infty) \) including the endpoints where the expression is zero.
5. **Final answer**:
\[
[0, 3) \cup (5, \infty)
\]
### (ii) Solve \( x^2 + 3x > 10 \)
1. **Rearrange the inequality**:
\[
x^2 + 3x - 10 > 0
\]
2. **Factor the quadratic**:
\[
(x + 5)(x - 2) > 0
\]
3. **Find the critical points**:
\( x = -5 \) and \( x = 2 \).
4. **Test intervals**: The intervals are \( (-\infty, -5) \), \( (-5, 2) \), and \( (2, \infty) \).
- For \( x = -6 \):
\(((-6 + 5)(-6 - 2) = (-1)(-8) > 0\)
- For \( x = 0 \):
\(((0 + 5)(0 - 2) = (5)(-2) < 0\)
- For \( x = 3 \):
\(((3 + 5)(3 - 2) = (8)(1) > 0\)
5. **Combine the results**: The solution is \( (-\infty, -5) \cup (2, \infty) \).
### Final answer for part (ii):
\[
(-\infty, -5) \cup (2, \infty)
\]
### (iii) Solve \( 9 - 2|4x + 1| > 3 \)
1. **Rearrange the inequality**:
\(-2|4x + 1| > -6\)
Dividing by -2 reverses the inequality:
\(|4x + 1| < 3\)
2. **Solve the absolute value inequality**:
\(-3 < 4x + 1 < 3\)
- For \(-3 < 4x + 1\):
\(-4 < 4x\)
\(x > -1\)
- For \(4x + 1 < 3\):
\(4x < 2\)
\(x < \frac{1}{2}\)
3. **Combine the results**:
\(-1 < x < \frac{1}{2}\)
### Final answer for part (iii):
\[
(-1, \frac{1}{2})
\]
### 2. Express \(-3 \leq x \leq 1\) in the form \(|x - a| \leq b\)
To express the inequality \(-3 \leq x \leq 1\), we look for an \(a\) and \(b\):
1. **Find the midpoint \(a\)**:
\[
a = \frac{-3 + 1}{2} = -1
\]
2. **Find half the distance (which is \(b\))**:
\[
b = \frac{1 - (-3)}{2} = \frac{4}{2} = 2
\]
3. **Final expression**:
\[
|x + 1| \leq 2
\]
Thus, you can express the given inequality as \(|x + 1| \leq 2\).
### (i) Solve \( \frac{x^2 - 5x}{x - 3} \geq 0 \)
1. **Factor the numerator**:
\[
x^2 - 5x = x(x - 5)
\]
So, we can rewrite the inequality as:
\[
\frac{x(x - 5)}{x - 3} \geq 0
\]
2. **Find the critical points**: Set the numerator and denominator equal to zero:
- \( x = 0 \) (from \( x \))
- \( x = 5 \) (from \( x - 5 \))
- \( x = 3 \) (from \( x - 3 \))
Thus, the critical points are \( x = 0, 3, 5 \).
3. **Test intervals**: The critical points divide the number line into the intervals: \( (-\infty, 0) \), \( (0, 3) \), \( (3, 5) \), and \( (5, \infty) \).
- Choose a test point from each interval:
- For \( x = -1 \) in \( (-\infty, 0) \):
\[
\frac{(-1)(-6)}{-4} = \frac{6}{-4} < 0
\]
- For \( x = 1 \) in \( (0, 3) \):
\[
\frac{(1)(-4)}{-2} = \frac{-4}{-2} > 0
\]
- For \( x = 4 \) in \( (3, 5) \):
\[
\frac{(4)(-1)}{1} = \frac{-4}{1} < 0
\]
- For \( x = 6 \) in \( (5, \infty) \):
\[
\frac{(6)(1)}{3} = \frac{6}{3} > 0
\]
4. **Combine the results**: The intervals where the expression is non-negative are \( [0, 3) \) and \( (5, \infty) \) including the endpoints where the expression is zero.
5. **Final answer**:
\[
[0, 3) \cup (5, \infty)
\]
### (ii) Solve \( x^2 + 3x > 10 \)
1. **Rearrange the inequality**:
\[
x^2 + 3x - 10 > 0
\]
2. **Factor the quadratic**:
\[
(x + 5)(x - 2) > 0
\]
3. **Find the critical points**:
\( x = -5 \) and \( x = 2 \).
4. **Test intervals**: The intervals are \( (-\infty, -5) \), \( (-5, 2) \), and \( (2, \infty) \).
- For \( x = -6 \):
\(((-6 + 5)(-6 - 2) = (-1)(-8) > 0\)
- For \( x = 0 \):
\(((0 + 5)(0 - 2) = (5)(-2) < 0\)
- For \( x = 3 \):
\(((3 + 5)(3 - 2) = (8)(1) > 0\)
5. **Combine the results**: The solution is \( (-\infty, -5) \cup (2, \infty) \).
### Final answer for part (ii):
\[
(-\infty, -5) \cup (2, \infty)
\]
### (iii) Solve \( 9 - 2|4x + 1| > 3 \)
1. **Rearrange the inequality**:
\(-2|4x + 1| > -6\)
Dividing by -2 reverses the inequality:
\(|4x + 1| < 3\)
2. **Solve the absolute value inequality**:
\(-3 < 4x + 1 < 3\)
- For \(-3 < 4x + 1\):
\(-4 < 4x\)
\(x > -1\)
- For \(4x + 1 < 3\):
\(4x < 2\)
\(x < \frac{1}{2}\)
3. **Combine the results**:
\(-1 < x < \frac{1}{2}\)
### Final answer for part (iii):
\[
(-1, \frac{1}{2})
\]
### 2. Express \(-3 \leq x \leq 1\) in the form \(|x - a| \leq b\)
To express the inequality \(-3 \leq x \leq 1\), we look for an \(a\) and \(b\):
1. **Find the midpoint \(a\)**:
\[
a = \frac{-3 + 1}{2} = -1
\]
2. **Find half the distance (which is \(b\))**:
\[
b = \frac{1 - (-3)}{2} = \frac{4}{2} = 2
\]
3. **Final expression**:
\[
|x + 1| \leq 2
\]
Thus, you can express the given inequality as \(|x + 1| \leq 2\).
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