`(3x^2+4x+3)^2-(k+1)(3x^2+4x+2)(3x^2+4x+3)+k(3x^2+4x+2)^2=0`. Find 'k' for which , equation has real roots.

Options

A. `(-1/2,1)`
B. `(1,5/2]`
C. `(1/2,1)`
D. `(-1,5/2]`

Answer

To solve the equation

\[ (3x^2 + 4x + 3)^2 - (k + 1)(3x^2 + 4x + 2)(3x^2 + 4x + 3) + k(3x^2 + 4x + 2)^2 = 0 \]

and find the values of \( k \) for which the equation has real roots, we can follow these steps:

### Step 1: Substitute \( T = 3x^2 + 4x + 2 \)

Let \( T = 3x^2 + 4x + 2 \). Then, we can express \( 3x^2 + 4x + 3 \) as \( T + 1 \).

### Step 2: Rewrite the equation

Substituting \( T \) into the equation gives us:

\[ (T + 1)^2 - (k + 1)T(T + 1) + kT^2 = 0 \]

### Step 3: Expand the equation

Expanding the left-hand side:

\[ T^2 + 2T + 1 - (k + 1)(T^2 + T) + kT^2 = 0 \]

This simplifies to:

\[ T^2 + 2T + 1 - (k + 1)T^2 - (k + 1)T + kT^2 = 0 \]

### Step 4: Combine like terms

Combining the terms gives:

\[ (1 - (k + 1) + k)T^2 + (2 - (k + 1))T + 1 = 0 \]

This simplifies to:

\[ (1 - 1)T^2 + (2 - k - 1)T + 1 = 0 \]

Thus, we have:

\[ (1 - k)T + 1 = 0 \]

### Step 5: Solve for \( T \)

Rearranging gives:

\[ T = \frac{-1}{1 - k} \]

### Step 6: Substitute back for \( T \)

Since \( T = 3x^2 + 4x + 2 \), we have:

\[ 3x^2 + 4x + 2 = \frac{-1}{1 - k} \]

### Step 7: Rearrange into standard form

This leads to:

\[ 3x^2 + 4x + \left(2 + \frac{1}{1 - k}\right) = 0 \]

### Step 8: Determine conditions for real roots

For the quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Here, \( a = 3 \), \( b = 4 \), and \( c = 2 + \frac{1}{1 - k} \).

Calculating the discriminant:

\[ D = 4^2 - 4 \cdot 3 \cdot \left(2 + \frac{1}{1 - k}\right) \geq 0 \]

This simplifies to:

\[ 16 - 12\left(2 + \frac{1}{1 - k}\right) \geq 0 \]

### Step 9: Solve the inequality

Expanding gives:

\[ 16 - 24 - \frac{12}{1 - k} \geq 0 \]

This leads to:

\[ -8 - \frac{12}{1 - k} \geq 0 \]

Rearranging gives:

\[ \frac{12}{1 - k} \leq -8 \]

Cross-multiplying (and flipping the inequality sign because \( 1 - k < 0 \)) gives:

\[ 12 \geq -8(1 - k) \]

### Step 10: Solve for \( k \)

Expanding and rearranging gives:

\[ 12 \geq -8 + 8k \]

Thus:

\[ 20 \geq 8k \quad \Rightarrow \quad k \leq \frac{20}{8} = \frac{5}{2} \]

### Step 11: Determine lower bound for \( k \)

Since \( 1 - k < 0 \), we also have \( k > 1 \).

### Final Result

Thus, the range of \( k \) for which the equation has real roots is:

\[ k \in (1, \frac{5}{2}] \]