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 1: Set up the equations
We start with the equations for \( P(x) \) and \( Q(x) \):
\[ P(x) = x^2 + \frac{1}{2}x + b \]
\[ Q(x) = x^2 + cx + d \]
The condition given is:
\[ P(x)Q(x) = Q(P(x)) \]

### Step 2: Expand both sides
First, we expand \( P(x)Q(x) \):
\[ P(x)Q(x) = (x^2 + \frac{1}{2}x + b)(x^2 + cx + d) \]
Expanding this product:
\[ = x^4 + cx^3 + dx^2 + \frac{1}{2}x^3 + \frac{1}{2}cx^2 + \frac{1}{2}dx + bx^2 + bcx + bd \]
Combining like terms:
\[ = x^4 + \left(c + \frac{1}{2}\right)x^3 + \left(d + \frac{1}{2}c + b\right)x^2 + \left(\frac{1}{2}d + bc\right)x + bd \]

Next, we expand \( Q(P(x)) \):
\[ Q(P(x)) = Q\left(x^2 + \frac{1}{2}x + b\right) = \left(x^2 + \frac{1}{2}x + b\right)^2 + c\left(x^2 + \frac{1}{2}x + b\right) + d \]
Expanding \( (x^2 + \frac{1}{2}x + b)^2 \):
\[ = x^4 + x^2 + b + \frac{1}{4}x^2 + cx^2 + \frac{1}{2}cx + cb + d \]
Combining like terms:
\[ = x^4 + \left(1 + c + \frac{1}{4}\right)x^2 + \left(\frac{1}{2}c\right)x + \left(b^2 + cb + d\right) \]

### Step 3: Set coefficients equal
Since \( P(x)Q(x) = Q(P(x)) \), we can equate the coefficients of \( x^4, x^3, x^2, x, \) and the constant term:
1. Coefficient of \( x^3 \): \( c + \frac{1}{2} = 0 \) → \( c = -\frac{1}{2} \)
2. Coefficient of \( x^2 \): \( d + \frac{1}{2}c + b = 1 + c + \frac{1}{4} \)
3. Coefficient of \( x \): \( \frac{1}{2}d + bc = \frac{1}{2}c \)
4. Constant term: \( bd = b^2 + cb + d \)

### Step 4: Solve for \( b \) and \( d \)
From \( c = -\frac{1}{2} \):
- Substitute \( c \) into the equations:
1. From \( d + \frac{1}{2}(-\frac{1}{2}) + b = 1 - \frac{1}{2} + \frac{1}{4} \):
\[ d - \frac{1}{4} + b = \frac{3}{4} \implies d + b = 1 \]
2. From \( \frac{1}{2}d + b(-\frac{1}{2}) = -\frac{1}{4} \):
\[ \frac{1}{2}d - \frac{1}{2}b = -\frac{1}{4} \implies d - b = -\frac{1}{2} \]

### Step 5: Solve the system of equations
We have:
1. \( d + b = 1 \)
2. \( d - b = -\frac{1}{2} \)

Adding these two equations:
\[ 2d = \frac{1}{2} \implies d = \frac{1}{4} \]
Substituting \( d \) back:
\[ \frac{1}{4} + b = 1 \implies b = \frac{3}{4} \]

### Step 6: Find the roots of \( P(Q(x)) = 0 \)
Now we have:
\[ P(x) = x^2 + \frac{1}{2}x + \frac{3}{4} \]
\[ Q(x) = x^2 - \frac{1}{2}x + \frac{1}{4} \]
To find \( P(Q(x)) = 0 \):
\[ P(Q(x)) = Q(x)^2 + \frac{1}{2}Q(x) + \frac{3}{4} \]
Substituting \( Q(x) \):
\[ = \left(x^2 - \frac{1}{2}x + \frac{1}{4}\right)^2 + \frac{1}{2}\left(x^2 - \frac{1}{2}x + \frac{1}{4}\right) + \frac{3}{4} \]
This is a quadratic equation in \( x \).

### Step 7: Solve the quadratic
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -\frac{1}{2}, c = \frac{3}{4} \).

### Final Step: Find real roots
After calculating the discriminant and solving, we find the real roots of the equation.