For what value of 'k' will the equation `2k x ^(2) -2 (1+2k)x + (3 +2k) =0` have real but distinct roots ? When will the roots be equal ?

Answer

To determine the value of 'k' for which the equation \(2k x^2 - 2(1 + 2k)x + (3 + 2k) = 0\) has real but distinct roots, we need to analyze the discriminant of the quadratic equation. The discriminant \(d\) is given by the formula:

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

### Step 1: Identify coefficients
In our equation, we can identify:
- \(a = 2k\)
- \(b = -2(1 + 2k)\)
- \(c = 3 + 2k\)

### Step 2: Calculate the discriminant
Now, we will calculate the discriminant \(d\):

\[ d = (-2(1 + 2k))^2 - 4(2k)(3 + 2k) \]

Calculating \(b^2\):

\[ b^2 = (-2(1 + 2k))^2 = 4(1 + 2k)^2 = 4(1 + 4k + 4k^2) = 4 + 16k + 16k^2 \]

Calculating \(4ac\):

\[ 4ac = 4(2k)(3 + 2k) = 8k(3 + 2k) = 24k + 16k^2 \]

### Step 3: Substitute in the discriminant formula
Now substituting \(b^2\) and \(4ac\) into the discriminant formula:

\[ d = (4 + 16k + 16k^2) - (24k + 16k^2) \]

### Step 4: Simplify the discriminant
Now, simplifying the expression:

\[ d = 4 + 16k + 16k^2 - 24k - 16k^2 \]
\[ d = 4 - 8k \]

### Step 5: Set the condition for real and distinct roots
For the roots to be real and distinct, we need:

\[ d > 0 \]
\[ 4 - 8k > 0 \]

### Step 6: Solve the inequality
Now, solving the inequality:

\[ 4 > 8k \]
\[ \frac{4}{8} > k \]
\[ \frac{1}{2} > k \]

Thus, we have:

\[ k < \frac{1}{2} \]

### Step 7: Condition for equal roots
For the roots to be equal, we set the discriminant equal to zero:

\[ d = 0 \]
\[ 4 - 8k = 0 \]

### Step 8: Solve for k
Solving for \(k\):

\[ 4 = 8k \]
\[ k = \frac{4}{8} = \frac{1}{2} \]

### Final Answer
Thus, the value of \(k\) for which the equation has real but distinct roots is \(k < \frac{1}{2}\), and the roots will be equal when \(k = \frac{1}{2}\).

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