Find the solution set of the equation `3x^(2)-8x-3=0`, when :
(i) `xepsilonZ( "integers" )`
(ii) `xepsilonQ( "rational numbers" )`.

Answer

To solve the quadratic equation \(3x^2 - 8x - 3 = 0\), we will follow these steps:

### Step 1: Write down the equation
We start with the equation:
\[ 3x^2 - 8x - 3 = 0 \]

### Step 2: Factor the quadratic equation
To factor the equation, we look for two numbers that multiply to \(3 \times -3 = -9\) and add up to \(-8\). The numbers \(-9\) and \(1\) work because:
\[ -9 + 1 = -8 \quad \text{and} \quad -9 \times 1 = -9 \]
Now we can rewrite the equation:
\[ 3x^2 - 9x + 1x - 3 = 0 \]

### Step 3: Group the terms
Next, we group the terms:
\[ (3x^2 - 9x) + (1x - 3) = 0 \]

### Step 4: Factor by grouping
Now we factor out the common factors from each group:
\[ 3x(x - 3) + 1(x - 3) = 0 \]
This can be factored further:
\[ (3x + 1)(x - 3) = 0 \]

### Step 5: Set each factor to zero
Now we set each factor equal to zero:
1. \(3x + 1 = 0\)
2. \(x - 3 = 0\)

### Step 6: Solve for \(x\)
From the first equation:
\[ 3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3} \]
From the second equation:
\[ x - 3 = 0 \implies x = 3 \]

### Step 7: Solution set for integers
Now, we need to find the solution set when \(x \in \mathbb{Z}\) (integers):
- The integer solutions are:
\[ \{3\} \]

### Step 8: Solution set for rational numbers
Next, we find the solution set when \(x \in \mathbb{Q}\) (rational numbers):
- The rational solutions are:
\[ \left\{3, -\frac{1}{3}\right\} \]

### Final Solution
Thus, the solution sets are:
(i) For \(x \in \mathbb{Z}\): \(\{3\}\)
(ii) For \(x \in \mathbb{Q}\): \(\{3, -\frac{1}{3}\}\)

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