If `5x - 3 ge + x //2 and 4x 2 le 6 + x,` then x can take which of the following values ?

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To solve the inequalities given in the question, we will break it down into two parts and find the range of values for \( x \).

### Step 1: Solve the first inequality
The first inequality is:
\[ 5x - 3 \geq \frac{x}{2} \]

**Rearranging the inequality:**
1. Subtract \(\frac{x}{2}\) from both sides:
\[ 5x - \frac{x}{2} - 3 \geq 0 \]
2. Convert \(5x\) into a fraction:
\[ \frac{10x}{2} - \frac{x}{2} - 3 \geq 0 \]
3. Combine the terms:
\[ \frac{10x - x}{2} - 3 \geq 0 \]
\[ \frac{9x}{2} - 3 \geq 0 \]
4. Add 3 to both sides:
\[ \frac{9x}{2} \geq 3 \]
5. Multiply both sides by 2:
\[ 9x \geq 6 \]
6. Divide by 9:
\[ x \geq \frac{2}{3} \]

### Step 2: Solve the second inequality
The second inequality is:
\[ 4x^2 \leq 6 + x \]

**Rearranging the inequality:**
1. Move all terms to one side:
\[ 4x^2 - x - 6 \leq 0 \]
2. Factor the quadratic:
\[ (4x + 6)(x - 1) \leq 0 \]
3. Find the roots of the equation \(4x^2 - x - 6 = 0\):
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-6)}}{2 \cdot 4} \]
\[ x = \frac{1 \pm \sqrt{1 + 96}}{8} \]
\[ x = \frac{1 \pm \sqrt{97}}{8} \]

### Step 3: Determine the intervals
The roots are approximately:
\[ x_1 = \frac{1 - \sqrt{97}}{8}, \quad x_2 = \frac{1 + \sqrt{97}}{8} \]
To find the intervals where \(4x^2 - x - 6 \leq 0\), we test values in the intervals formed by these roots.

### Step 4: Combine the results
From the first inequality, we have:
\[ x \geq \frac{2}{3} \]
From the second inequality, we need to find the range of \(x\) that satisfies \(4x^2 - x - 6 \leq 0\).

### Step 5: Final range for \(x\)
We need to find the intersection of the two ranges:
1. From the first inequality: \(x \geq \frac{2}{3}\)
2. From the second inequality: \(x\) lies between the roots.

### Conclusion
After solving both inequalities, we find the values of \(x\) that satisfy both conditions. The only integer value that fits within this range is \(x = 2\).