Find the values of k for which roots of the following equations are real and equal:
(i) `12x^(2)+4kx+3=0` (ii) `kx^(2)-5x+k=0`
(iii) `x^(2)+k(4x+k-1)+2=0`

Answer

To find the values of \( k \) for which the roots of the given quadratic equations are real and equal, we need to set the discriminant \( \Delta \) of each equation to zero. The discriminant is given by:

\[ \Delta = b^2 - 4ac \]

where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \).

### (i) For the equation \( 12x^2 + 4kx + 3 = 0 \):

1. Identify the coefficients:
- \( a = 12 \)
- \( b = 4k \)
- \( c = 3 \)

2. Set up the discriminant:
\[ \Delta = (4k)^2 - 4 \cdot 12 \cdot 3 \]

3. Calculate \( \Delta \):
\[ \Delta = 16k^2 - 144 \]

4. Set the discriminant to zero for real and equal roots:
\[ 16k^2 - 144 = 0 \]

5. Solve for \( k \):
\[ 16k^2 = 144 \]
\[ k^2 = \frac{144}{16} = 9 \]
\[ k = \pm 3 \]

### (ii) For the equation \( kx^2 - 5x + k = 0 \):

1. Identify the coefficients:
- \( a = k \)
- \( b = -5 \)
- \( c = k \)

2. Set up the discriminant:
\[ \Delta = (-5)^2 - 4 \cdot k \cdot k \]

3. Calculate \( \Delta \):
\[ \Delta = 25 - 4k^2 \]

4. Set the discriminant to zero:
\[ 25 - 4k^2 = 0 \]

5. Solve for \( k \):
\[ 4k^2 = 25 \]
\[ k^2 = \frac{25}{4} \]
\[ k = \pm \frac{5}{2} \]

### (iii) For the equation \( x^2 + k(4x + k - 1) + 2 = 0 \):

1. Expand the equation:
\[ x^2 + 4kx + k^2 - k + 2 = 0 \]

2. Identify the coefficients:
- \( a = 1 \)
- \( b = 4k \)
- \( c = k^2 - k + 2 \)

3. Set up the discriminant:
\[ \Delta = (4k)^2 - 4 \cdot 1 \cdot (k^2 - k + 2) \]

4. Calculate \( \Delta \):
\[ \Delta = 16k^2 - 4(k^2 - k + 2) \]
\[ = 16k^2 - 4k^2 + 4k - 8 \]
\[ = 12k^2 + 4k - 8 \]

5. Set the discriminant to zero:
\[ 12k^2 + 4k - 8 = 0 \]

6. Simplify the equation:
\[ 3k^2 + k - 2 = 0 \]

7. Solve for \( k \) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
\[ = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6} \]
\[ k = \frac{4}{6} = \frac{2}{3} \quad \text{or} \quad k = \frac{-6}{6} = -1 \]

### Final Answers:
1. For \( 12x^2 + 4kx + 3 = 0 \), \( k = \pm 3 \)
2. For \( kx^2 - 5x + k = 0 \), \( k = \pm \frac{5}{2} \)
3. For \( x^2 + k(4x + k - 1) + 2 = 0 \), \( k = -1 \) or \( k = \frac{2}{3} \)