If `x^(2) - 3x + 2` is a factor of `x^(4) - ax^(2) + b` then the equation whose roots are a and b is

Options

A. `x^(2) - 9x - 20 = 0`
B. `x^(2) - 9x + 20 = 0`
C. `x^(2) + 9x - 20 = 0`
D. `x^(2) + 9x + 20 = 0`

Answer

To solve the problem step by step, we need to determine the values of \( a \) and \( b \) such that \( x^2 - 3x + 2 \) is a factor of \( x^4 - ax^2 + b \).

### Step 1: Identify the factor
The quadratic \( x^2 - 3x + 2 \) can be factored as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
This means that \( x = 1 \) and \( x = 2 \) are the roots of the quadratic.

### Step 2: Use the roots to form equations
Since \( x^2 - 3x + 2 \) is a factor of \( x^4 - ax^2 + b \), both \( x = 1 \) and \( x = 2 \) must satisfy the equation \( x^4 - ax^2 + b = 0 \).

#### For \( x = 1 \):
Substituting \( x = 1 \):
\[ 1^4 - a(1^2) + b = 0 \implies 1 - a + b = 0 \quad \text{(Equation 1)} \]

#### For \( x = 2 \):
Substituting \( x = 2 \):
\[ 2^4 - a(2^2) + b = 0 \implies 16 - 4a + b = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 1 - a + b = 0 \)
2. \( 16 - 4a + b = 0 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = a - 1 \quad \text{(Substituting into Equation 2)} \]

Substituting \( b = a - 1 \) into Equation 2:
\[ 16 - 4a + (a - 1) = 0 \]
Simplifying this:
\[ 16 - 4a + a - 1 = 0 \implies 15 - 3a = 0 \]
Thus,
\[ 3a = 15 \implies a = 5 \]

### Step 4: Find \( b \)
Now substituting \( a = 5 \) back into the equation for \( b \):
\[ b = 5 - 1 = 4 \]

### Step 5: Form the quadratic equation
Now we have \( a = 5 \) and \( b = 4 \). The equation whose roots are \( a \) and \( b \) can be written as:
\[ x^2 - (a + b)x + ab = 0 \]
Substituting \( a \) and \( b \):
\[ x^2 - (5 + 4)x + (5 \cdot 4) = 0 \]
This simplifies to:
\[ x^2 - 9x + 20 = 0 \]

### Final Answer
The equation whose roots are \( a \) and \( b \) is:
\[ \boxed{x^2 - 9x + 20 = 0} \]