Set `A={x:x inI,x^(4)-x^(3)-2x^(2)+2x=0}`
`B={x:x inN,2x^(2)-1lt7}`
Are A and B comparable ?

Answer

To determine if sets A and B are comparable, we need to find the elements of each set and then check if one set is a subset of the other or if they are equal.

### Step 1: Find the elements of Set A

Set A is defined as:
\[ A = \{ x : x \in \mathbb{I}, x^4 - x^3 - 2x^2 + 2x = 0 \} \]

To find the elements of Set A, we need to solve the equation:
\[ x^4 - x^3 - 2x^2 + 2x = 0 \]

First, we can factor out \( x \):
\[ x(x^3 - x^2 - 2x + 2) = 0 \]

This gives us one solution:
\[ x = 0 \]

Next, we need to solve the cubic equation:
\[ x^3 - x^2 - 2x + 2 = 0 \]

We can check for rational roots using the Rational Root Theorem. Testing \( x = 1 \):
\[ 1^3 - 1^2 - 2(1) + 2 = 1 - 1 - 2 + 2 = 0 \]
So, \( x = 1 \) is a root.

Now we can factor the cubic equation using synthetic division or polynomial long division:
\[ x^3 - x^2 - 2x + 2 = (x - 1)(x^2 + 2) \]

The quadratic \( x^2 + 2 = 0 \) has no real solutions, but we can find complex solutions:
\[ x^2 = -2 \]
\[ x = \pm \sqrt{2}i \]

Thus, the real solutions to the original equation are:
- \( x = 0 \)
- \( x = 1 \)

Therefore, Set A in roster form is:
\[ A = \{ 0, 1 \} \]

### Step 2: Find the elements of Set B

Set B is defined as:
\[ B = \{ x : x \in \mathbb{N}, 2x^2 - 1 < 7 \} \]

We can simplify the inequality:
\[ 2x^2 < 8 \]
\[ x^2 < 4 \]
\[ -2 < x < 2 \]

Since \( x \) must be a natural number (positive integers), the possible values for \( x \) are:
\[ x = 1 \]

Thus, Set B in roster form is:
\[ B = \{ 1 \} \]

### Step 3: Check if sets A and B are comparable

To determine if sets A and B are comparable, we check if:
1. \( A \subseteq B \)
2. \( B \subseteq A \)
3. \( A = B \)

From our findings:
- Set A: \( \{ 0, 1 \} \)
- Set B: \( \{ 1 \} \)

We see that:
- \( B \subseteq A \) since all elements of B (which is just 1) are also in A.

Since \( B \) is a subset of \( A \), we conclude that sets A and B are comparable.

### Final Answer
Yes, sets A and B are comparable.

---