Consider an unknow polynomial which divided by `(x - 3)` and `(x-4)` leaves remainder 2 and 1, respectively. Let R(x) be the remainder when this polynomial is divided by `(x-3)(x-4)`.
If equations `R(x) = x^(2)+ ax +1` has two distint real roots, then exhaustive values of a are.

Options

A. `(-2,2)`
B. `(-oo,-2) uu(2,oo)`
C. `(-2,oo)`
D. all real numbers

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Define the Polynomial and Remainder
Let \( P(x) \) be the unknown polynomial. According to the problem, when \( P(x) \) is divided by \( (x - 3)(x - 4) \), it leaves a remainder \( R(x) \). Since the degree of the remainder must be less than the degree of the divisor, we can express \( R(x) \) as a linear polynomial:
\[ R(x) = ax + b \]

### Step 2: Use the Remainder Theorem
We know that:
- \( P(3) = 2 \) (the remainder when divided by \( x - 3 \))
- \( P(4) = 1 \) (the remainder when divided by \( x - 4 \))

Thus, we can write:
\[ R(3) = 3a + b = 2 \quad \text{(1)} \]
\[ R(4) = 4a + b = 1 \quad \text{(2)} \]

### Step 3: Solve the System of Equations
Now we have a system of equations. We can subtract equation (1) from equation (2):
\[ (4a + b) - (3a + b) = 1 - 2 \]
This simplifies to:
\[ 4a - 3a = -1 \]
\[ a = -1 \]

### Step 4: Substitute Back to Find \( b \)
Now substitute \( a = -1 \) back into equation (1):
\[ 3(-1) + b = 2 \]
\[ -3 + b = 2 \]
\[ b = 5 \]

Thus, we have:
\[ R(x) = -x + 5 \]

### Step 5: Set Up the Quadratic Equation
We need to find the values of \( a \) such that the equation \( R(x) = x^2 + ax + 1 \) has two distinct real roots. This gives us:
\[ -x + 5 = x^2 + ax + 1 \]
Rearranging this, we get:
\[ x^2 + (a + 1)x - 4 = 0 \]

### Step 6: Use the Discriminant Condition
For the quadratic equation to have two distinct real roots, the discriminant must be greater than zero:
\[ D = (a + 1)^2 - 4(1)(-4) > 0 \]
\[ D = (a + 1)^2 + 16 > 0 \]

Since \( (a + 1)^2 \) is always non-negative, and adding 16 ensures that the discriminant is always positive. Therefore, the condition is satisfied for all real values of \( a \).

### Conclusion
The exhaustive values of \( a \) are:
\[ \text{All real numbers} \]