(i) Find the value of k for which x=1 is a root of the equation `x^(2)+kx+3=0.` Also, find the other root.
(ii) Find the values of a and b for which `x=(3)/(4)" and "x=-2` are the roots of the equation `ax^(2)+bx-6=0.`

Answer

Let's solve the given problem step by step.

### Part (i)
**Find the value of k for which x = 1 is a root of the equation \(x^2 + kx + 3 = 0\). Also, find the other root.**

1. **Substitute the root into the equation:**
Since \(x = 1\) is a root, we can substitute \(x = 1\) into the equation:
\[ 1^2 + k(1) + 3 = 0 \]
This simplifies to:
\[ 1 + k + 3 = 0 \]
\[ k + 4 = 0 \]

2. **Solve for k:**
Rearranging the equation gives:
\[ k = -4 \]

3. **Write the quadratic equation with k:**
Substitute \(k = -4\) back into the original equation:
\[ x^2 - 4x + 3 = 0 \]

4. **Factor the quadratic equation:**
We can factor this equation:
\[ (x - 1)(x - 3) = 0 \]

5. **Find the other root:**
The roots of the equation are \(x = 1\) and \(x = 3\). Thus, the other root is:
\[ x = 3 \]

### Summary for Part (i):
- The value of \(k\) is \(-4\).
- The other root is \(3\).

---

### Part (ii)
**Find the values of a and b for which \(x = \frac{3}{4}\) and \(x = -2\) are the roots of the equation \(ax^2 + bx - 6 = 0\).**

1. **Use the first root \(x = \frac{3}{4}\):**
Substitute \(x = \frac{3}{4}\) into the equation:
\[ a\left(\frac{3}{4}\right)^2 + b\left(\frac{3}{4}\right) - 6 = 0 \]
This simplifies to:
\[ a\left(\frac{9}{16}\right) + b\left(\frac{3}{4}\right) - 6 = 0 \]
Multiply through by \(16\) to eliminate the fraction:
\[ 9a + 12b - 96 = 0 \quad \text{(Equation 1)} \]

2. **Use the second root \(x = -2\):**
Substitute \(x = -2\) into the equation:
\[ a(-2)^2 + b(-2) - 6 = 0 \]
This simplifies to:
\[ 4a - 2b - 6 = 0 \quad \text{(Equation 2)} \]

3. **Rewrite Equation 2:**
Rearranging gives:
\[ 4a - 2b = 6 \quad \Rightarrow \quad 2a - b = 3 \quad \text{(Equation 2')} \]

4. **Solve the system of equations:**
Now we have two equations:
- \(9a + 12b = 96\) (Equation 1)
- \(2a - b = 3\) (Equation 2')

From Equation 2', we can express \(b\) in terms of \(a\):
\[ b = 2a - 3 \]

5. **Substitute \(b\) into Equation 1:**
Substitute \(b\) into Equation 1:
\[ 9a + 12(2a - 3) = 96 \]
This simplifies to:
\[ 9a + 24a - 36 = 96 \]
\[ 33a - 36 = 96 \]
\[ 33a = 132 \quad \Rightarrow \quad a = 4 \]

6. **Find \(b\):**
Substitute \(a = 4\) back into \(b = 2a - 3\):
\[ b = 2(4) - 3 = 8 - 3 = 5 \]

### Summary for Part (ii):
- The values are \(a = 4\) and \(b = 5\).

---