Let `P(x) =x^(2)+ 1/2x + b` and `Q(x) = x^(2) + cx + d` be two polynomials with real coefficients such that `P(x) Q(x) = Q(P(x))` for all real x. Find all the real roots of P(Q(x)) =0.

Answer

To solve the problem, we need to find the real roots of the equation \( P(Q(x)) = 0 \), given the polynomials \( P(x) = x^2 + \frac{1}{2}x + b \) and \( Q(x) = x^2 + cx + d \) such that \( P(x)Q(x) = Q(P(x)) \) for all real \( x \).

### Step-by-Step Solution:

1. **Set Up the Equation**:
We start with the equation:
\[ P(x)Q(x) = Q(P(x)) \]
Substituting the expressions for \( P(x) \) and \( Q(x) \):
\[ \left( x^2 + \frac{1}{2}x + b \right) \left( x^2 + cx + d \right) = Q\left( x^2 + \frac{1}{2}x + b \right) \]

2. **Expand Both Sides**:
Let's expand the left-hand side:
\[ P(x)Q(x) = \left( x^2 + \frac{1}{2}x + b \right) \left( x^2 + cx + d \right) \]
This results in:
\[ x^4 + cx^3 + dx^2 + \frac{1}{2}x^3 + \frac{c}{2}x^2 + \frac{d}{2}x + bx^2 + bcx + bd \]
Combining like terms gives:
\[ x^4 + \left(c + \frac{1}{2}\right)x^3 + \left(d + \frac{c}{2} + b\right)x^2 + \left(\frac{d}{2} + bc\right)x + bd \]

3. **Evaluate \( Q(P(x)) \)**:
Now we need to evaluate \( Q(P(x)) \):
\[ Q(P(x)) = Q\left(x^2 + \frac{1}{2}x + b\right) = \left(x^2 + \frac{1}{2}x + b\right)^2 + c\left(x^2 + \frac{1}{2}x + b\right) + d \]
Expanding \( \left(x^2 + \frac{1}{2}x + b\right)^2 \):
\[ = x^4 + x^3 + \left(\frac{1}{4} + 2b\right)x^2 + b^2 + c\left(x^2 + \frac{1}{2}x + b\right) + d \]
Combining terms gives:
\[ x^4 + x^3 + \left(\frac{1}{4} + 2b + c\right)x^2 + \left(\frac{c}{2}\right)x + (b^2 + cb + d) \]

4. **Equate Coefficients**:
Since \( P(x)Q(x) = Q(P(x)) \), we equate coefficients:
- Coefficient of \( x^4 \): \( 1 = 1 \) (satisfied)
- Coefficient of \( x^3 \): \( c + \frac{1}{2} = 1 \) → \( c = \frac{1}{2} \)
- Coefficient of \( x^2 \): \( d + \frac{1}{4} + b = \frac{1}{4} + 2b + \frac{1}{2} \)
- Coefficient of \( x \): \( \frac{d}{2} + bc = \frac{c}{2} \)
- Constant term: \( bd = b^2 + cb + d \)

5. **Solve for \( b \) and \( d \)**:
From the coefficient equations, we can derive values for \( b \) and \( d \):
- From \( d + b = 2b + \frac{1}{2} \): \( d = b + \frac{1}{2} \)
- From \( \frac{d}{2} + \frac{1}{2}b = \frac{1}{4} \): substituting \( d \) gives us a quadratic equation in \( b \).

6. **Find Roots of \( P(Q(x)) = 0 \)**:
Substitute \( b \) and \( d \) back into \( P(x) \) and \( Q(x) \):
\[ P(x) = x^2 + \frac{1}{2}x - \frac{1}{2}, \quad Q(x) = x^2 + \frac{1}{2}x \]
Now, we need to find the roots of \( P(Q(x)) = 0 \).

7. **Calculate the Roots**:
Set \( P(Q(x)) = 0 \):
\[ Q(x) = 0 \Rightarrow x(x + \frac{1}{2}) = 0 \Rightarrow x = 0 \text{ or } x = -\frac{1}{2} \]
Substitute these values into \( P(x) \) to find the corresponding roots.

### Final Answer:
The real roots of \( P(Q(x)) = 0 \) are:
\[ x = 0 \quad \text{and} \quad x = -\frac{1}{2} \]