If the equation x^4 -λx^2 +9 has only two real roods, then the set of values of `λ` is

Options

A. `(-oo,-6)`
B. `(-6,6)`
C. `{6}`
D. `phi`

Explanation

To solve the problem, we need to analyze the given polynomial equation \( x^4 - \lambda x^2 + 9 = 0 \) and determine the conditions under which it has exactly two real roots.

### Step-by-Step Solution:

1. **Rewrite the Equation**:
We start with the equation:
\[ x^4 - \lambda x^2 + 9 = 0 \]
Let's substitute \( y = x^2 \). Then the equation becomes:
\[ y^2 - \lambda y + 9 = 0 \]

2. **Determine the Nature of Roots**:
The roots of the quadratic equation \( y^2 - \lambda y + 9 = 0 \) can be found using the discriminant \( D \):
\[ D = b^2 - 4ac = (-\lambda)^2 - 4 \cdot 1 \cdot 9 = \lambda^2 - 36 \]

3. **Conditions for Roots**:
For the original quartic equation to have exactly two real roots, the quadratic in \( y \) must have one double root (which corresponds to two equal roots in \( x \)), or two distinct roots that yield only two real values for \( x \).

- **Double Root Condition**:
For the quadratic to have a double root:
\[ D = 0 \implies \lambda^2 - 36 = 0 \implies \lambda^2 = 36 \implies \lambda = 6 \text{ or } \lambda = -6 \]

- **Two Distinct Roots Condition**:
For the quadratic to have two distinct roots, we need:
\[ D > 0 \implies \lambda^2 - 36 > 0 \implies \lambda^2 > 36 \implies \lambda < -6 \text{ or } \lambda > 6 \]

4. **Combining Conditions**:
The conditions for having exactly two real roots (either as a double root or two distinct roots) can be summarized as:
- \( \lambda = 6 \) (double root)
- \( \lambda = -6 \) (double root)
- \( \lambda < -6 \) (two distinct roots)
- \( \lambda > 6 \) (two distinct roots)

5. **Conclusion**:
The set of values of \( \lambda \) for which the equation \( x^4 - \lambda x^2 + 9 = 0 \) has exactly two real roots is:
\[ (-\infty, -6) \cup (6, \infty) \]
Therefore, the answer is that the set of values of \( \lambda \) is \( (-\infty, -6) \cup (6, \infty) \).