The quadratic equation `x^(2) + (a^(2) - 2) x - 2a^(2) and x^(2) - 3x + 2 = 0` have

Options

A. no common root for all `a in R`
B. exactly one common root for all `a in R`
C. two common roots for some `a in R`
D. none of these

Answer

To solve the problem, we need to analyze the two quadratic equations given:

1. \( x^2 + (a^2 - 2)x - 2a^2 = 0 \) (Equation 1)
2. \( x^2 - 3x + 2 = 0 \) (Equation 2)

### Step 1: Solve Equation 2

First, let's solve Equation 2:

\[ x^2 - 3x + 2 = 0 \]

We can factor this equation:

\[ (x - 1)(x - 2) = 0 \]

This gives us the roots:

\[ x = 1 \quad \text{and} \quad x = 2 \]

### Step 2: Check for Common Roots

Next, we need to check if either of these roots (1 or 2) is also a root of Equation 1.

#### Check for \( x = 1 \):

Substituting \( x = 1 \) into Equation 1:

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

Simplifying this:

\[ 1 + a^2 - 2 - 2a^2 = 0 \]
\[ -a^2 - 1 = 0 \]
\[ -a^2 = 1 \]

This equation cannot hold true for any real number \( a \) since \( a^2 \) is always non-negative. Therefore, \( x = 1 \) is not a common root.

#### Check for \( x = 2 \):

Now, substituting \( x = 2 \) into Equation 1:

\[ 2^2 + (a^2 - 2)(2) - 2a^2 = 0 \]

Simplifying this:

\[ 4 + 2a^2 - 4 - 2a^2 = 0 \]
\[ 0 = 0 \]

This is always true, which means \( x = 2 \) is a common root for all values of \( a \).

### Conclusion

Thus, the two quadratic equations have exactly one common root, which is \( x = 2 \).

The correct answer is:

**Option B: Exactly one common root for all \( a \) belonging to real numbers.**