If the two roots of the euqation `(lamda-1) (x^(2)+x+1)^(2)-(lamda+1)(x^(4)+x^(2)+1)=0` are real and distinct, then `lamda` lies in the interval

Options

A. `(-oo,2)`
B. `(2,oo)`
C. `(-oo,-2)`
D. `(-oo,-2) cap (2,oo)`

Answer

To solve the problem, we need to analyze the equation given and determine the conditions under which the roots are real and distinct. The equation is:

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

### Step 1: Simplify the Equation

We can factor out \( (x^2 + x + 1) \) from both terms:

\[ (x^2 + x + 1) \left[ (\lambda - 1)(x^2 + x + 1) - (\lambda + 1)(x^2 - x + 1) \right] = 0 \]

### Step 2: Set Up the Quadratic Equation

The equation can be simplified to:

\[ (\lambda - 1)(x^2 + x + 1) - (\lambda + 1)(x^2 - x + 1) = 0 \]

Expanding this gives:

\[ (\lambda - 1)(x^2 + x + 1) = (\lambda + 1)(x^2 - x + 1) \]

### Step 3: Collect Like Terms

Rearranging the equation leads to:

\[ (\lambda - 1)x^2 + (\lambda - 1)x + (\lambda - 1) = (\lambda + 1)x^2 - (\lambda + 1)x + (\lambda + 1) \]

Combining like terms results in:

\[ (\lambda - 1 - \lambda - 1)x^2 + (\lambda - 1 + \lambda + 1)x + (\lambda - 1 - \lambda - 1) = 0 \]

This simplifies to:

\[ (-2)x^2 + (2\lambda)x + (-2) = 0 \]

### Step 4: Formulate the Quadratic Equation

We can factor out \(-2\):

\[ -2(x^2 - \lambda x + 1) = 0 \]

Thus, we have the quadratic equation:

\[ x^2 - \lambda x + 1 = 0 \]

### Step 5: Determine Conditions for Real and Distinct Roots

For the roots of the quadratic equation to be real and distinct, the discriminant must be greater than zero:

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

Here, \(a = 1\), \(b = -\lambda\), and \(c = 1\):

\[ (-\lambda)^2 - 4(1)(1) > 0 \]

This simplifies to:

\[ \lambda^2 - 4 > 0 \]

### Step 6: Solve the Inequality

Factoring the inequality gives:

\[ (\lambda - 2)(\lambda + 2) > 0 \]

### Step 7: Analyze the Intervals

To find the intervals where this inequality holds, we identify the critical points:

- \(\lambda = -2\)
- \(\lambda = 2\)

Using a number line, we test the intervals:

1. For \(\lambda < -2\): Choose \(\lambda = -3\) → Positive
2. For \(-2 < \lambda < 2\): Choose \(\lambda = 0\) → Negative
3. For \(\lambda > 2\): Choose \(\lambda = 3\) → Positive

Thus, the solution to the inequality is:

\[ \lambda \in (-\infty, -2) \cup (2, \infty) \]

### Final Answer

The values of \(\lambda\) for which the roots of the equation are real and distinct are:

\[ \lambda \in (-\infty, -2) \cup (2, \infty) \]