Answer :
To solve the inequality [tex]\( 2x + \frac{3}{x} \leq 5 \)[/tex], follow these steps while providing a detailed solution:
1. Rearrange the inequality:
We start with the inequality:
[tex]\[ 2x + \frac{3}{x} \leq 5 \][/tex]
2. Move all terms to one side:
Subtract 5 from both sides to set the inequality to zero:
[tex]\[ 2x + \frac{3}{x} - 5 \leq 0 \][/tex]
3. Find a common denominator:
To simplify the expression, find a common denominator which is [tex]\( x \)[/tex]:
[tex]\[ \frac{2x^2}{x} + \frac{3}{x} - \frac{5x}{x} \leq 0 \][/tex]
Combine the terms:
[tex]\[ \frac{2x^2 + 3 - 5x}{x} \leq 0 \][/tex]
4. Solve the quadratic equation in the numerator:
Consider the quadratic equation in the numerator:
[tex]\[ 2x^2 - 5x + 3 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4} \][/tex]
Thus, we get the roots:
[tex]\[ x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}, \quad x = \frac{5 - 1}{4} = \frac{4}{4} = 1 \][/tex]
5. Identify the critical points and intervals:
The critical points [tex]\( x = 1 \)[/tex] and [tex]\( x = \frac{3}{2} \)[/tex] divide the number line into the intervals:
[tex]\[ (-\infty, 0), \quad (0, 1), \quad (1, \frac{3}{2}), \quad (\frac{3}{2}, \infty) \][/tex]
Note that [tex]\( x \)[/tex] must be non-zero since it's in the denominator.
6. Test the intervals:
We need to determine where the expression [tex]\( \frac{2x^2 - 5x + 3}{x} \)[/tex] is less than or equal to zero:
- For [tex]\( x \in (-\infty, 0) \)[/tex]:
Substitute a test point such as [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{2(-1)^2 - 5(-1) + 3}{-1} = \frac{2 + 5 + 3}{-1} = \frac{10}{-1} = -10 \quad (\text{Negative}) \][/tex]
The interval [tex]\( (-\infty, 0) \)[/tex] satisfies the inequality.
- For [tex]\( x \in (0, 1) \)[/tex]:
Substitute a test point such as [tex]\( x = 0.5 \)[/tex]:
[tex]\[ \frac{2(0.5)^2 - 5(0.5) + 3}{0.5} = \frac{0.5 - 2.5 + 3}{0.5} = \frac{1}{0.5} = 2 \quad (\text{Not negative}) \][/tex]
The interval [tex]\( (0, 1) \)[/tex] does not satisfy the inequality.
- For [tex]\( x \in (1, \frac{3}{2}) \)[/tex]:
Substitute a test point such as [tex]\( x = 1.25 \)[/tex]:
[tex]\[ \frac{2(1.25)^2 - 5(1.25) + 3}{1.25} = \frac{3.125 - 6.25 + 3}{1.25} = \frac{-0.125}{1.25} = -0.1 \quad (\text{Negative}) \][/tex]
The interval [tex]\( (1, \frac{3}{2}) \)[/tex] satisfies the inequality.
- For [tex]\( x \in (\frac{3}{2}, \infty) \)[/tex]:
Substitute a test point such as [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{2(2)^2 - 5(2) + 3}{2} = \frac{8 - 10 + 3}{2} = \frac{1}{2} = 0.5 \quad (\text{Not negative}) \][/tex]
The interval [tex]\( (\frac{3}{2}, \infty) \)[/tex] does not satisfy the inequality.
7. Combine the valid intervals:
The intervals that satisfy the inequality are:
[tex]\[ (-\infty, 0) \quad \text{and} \quad (1, \frac{3}{2}) \][/tex]
Including the endpoints where the expression evaluates to zero:
[tex]\[ x = 1 \quad \text{and} \quad x = \frac{3}{2} \][/tex]
Therefore, the solution in interval notation is:
[tex]\[ (-\infty, 0) \cup [1, \frac{3}{2}] \][/tex]
Thus, the solution to the inequality [tex]\( 2x + \frac{3}{x} \leq 5 \)[/tex] is:
[tex]\[ x \in (-\infty, 0) \cup [1, \frac{3}{2}] \][/tex]
1. Rearrange the inequality:
We start with the inequality:
[tex]\[ 2x + \frac{3}{x} \leq 5 \][/tex]
2. Move all terms to one side:
Subtract 5 from both sides to set the inequality to zero:
[tex]\[ 2x + \frac{3}{x} - 5 \leq 0 \][/tex]
3. Find a common denominator:
To simplify the expression, find a common denominator which is [tex]\( x \)[/tex]:
[tex]\[ \frac{2x^2}{x} + \frac{3}{x} - \frac{5x}{x} \leq 0 \][/tex]
Combine the terms:
[tex]\[ \frac{2x^2 + 3 - 5x}{x} \leq 0 \][/tex]
4. Solve the quadratic equation in the numerator:
Consider the quadratic equation in the numerator:
[tex]\[ 2x^2 - 5x + 3 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4} \][/tex]
Thus, we get the roots:
[tex]\[ x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}, \quad x = \frac{5 - 1}{4} = \frac{4}{4} = 1 \][/tex]
5. Identify the critical points and intervals:
The critical points [tex]\( x = 1 \)[/tex] and [tex]\( x = \frac{3}{2} \)[/tex] divide the number line into the intervals:
[tex]\[ (-\infty, 0), \quad (0, 1), \quad (1, \frac{3}{2}), \quad (\frac{3}{2}, \infty) \][/tex]
Note that [tex]\( x \)[/tex] must be non-zero since it's in the denominator.
6. Test the intervals:
We need to determine where the expression [tex]\( \frac{2x^2 - 5x + 3}{x} \)[/tex] is less than or equal to zero:
- For [tex]\( x \in (-\infty, 0) \)[/tex]:
Substitute a test point such as [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{2(-1)^2 - 5(-1) + 3}{-1} = \frac{2 + 5 + 3}{-1} = \frac{10}{-1} = -10 \quad (\text{Negative}) \][/tex]
The interval [tex]\( (-\infty, 0) \)[/tex] satisfies the inequality.
- For [tex]\( x \in (0, 1) \)[/tex]:
Substitute a test point such as [tex]\( x = 0.5 \)[/tex]:
[tex]\[ \frac{2(0.5)^2 - 5(0.5) + 3}{0.5} = \frac{0.5 - 2.5 + 3}{0.5} = \frac{1}{0.5} = 2 \quad (\text{Not negative}) \][/tex]
The interval [tex]\( (0, 1) \)[/tex] does not satisfy the inequality.
- For [tex]\( x \in (1, \frac{3}{2}) \)[/tex]:
Substitute a test point such as [tex]\( x = 1.25 \)[/tex]:
[tex]\[ \frac{2(1.25)^2 - 5(1.25) + 3}{1.25} = \frac{3.125 - 6.25 + 3}{1.25} = \frac{-0.125}{1.25} = -0.1 \quad (\text{Negative}) \][/tex]
The interval [tex]\( (1, \frac{3}{2}) \)[/tex] satisfies the inequality.
- For [tex]\( x \in (\frac{3}{2}, \infty) \)[/tex]:
Substitute a test point such as [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{2(2)^2 - 5(2) + 3}{2} = \frac{8 - 10 + 3}{2} = \frac{1}{2} = 0.5 \quad (\text{Not negative}) \][/tex]
The interval [tex]\( (\frac{3}{2}, \infty) \)[/tex] does not satisfy the inequality.
7. Combine the valid intervals:
The intervals that satisfy the inequality are:
[tex]\[ (-\infty, 0) \quad \text{and} \quad (1, \frac{3}{2}) \][/tex]
Including the endpoints where the expression evaluates to zero:
[tex]\[ x = 1 \quad \text{and} \quad x = \frac{3}{2} \][/tex]
Therefore, the solution in interval notation is:
[tex]\[ (-\infty, 0) \cup [1, \frac{3}{2}] \][/tex]
Thus, the solution to the inequality [tex]\( 2x + \frac{3}{x} \leq 5 \)[/tex] is:
[tex]\[ x \in (-\infty, 0) \cup [1, \frac{3}{2}] \][/tex]
