Let `(a_(1),a_(2),a_(3),a_(4),a_(5))` denote a re=arrangement of (3,-5,7,4-9), then `a_(1)x^(4)+a_(2)x^(3)+a_(3)x^(2)+a_(4)+a_(5)=0` has

Options

A. at least two real roots
B. all four real roots
C. only imaginary roots
D. None of these

Explanation

To solve the given problem, we need to analyze the polynomial formed by the rearrangement of the coefficients and determine the nature of its roots.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficients \( (a_1, a_2, a_3, a_4, a_5) \) are a rearrangement of the numbers \( (3, -5, 7, 4, -9) \).

2. **Form the Polynomial**:
The polynomial can be expressed as:
\[ f(x) = a_1 x^4 + a_2 x^3 + a_3 x^2 + a_4 x + a_5 = 0 \]

3. **Evaluate at \( x = 1 \)**:
We substitute \( x = 1 \) into the polynomial:
\[ f(1) = a_1(1^4) + a_2(1^3) + a_3(1^2) + a_4(1) + a_5 = a_1 + a_2 + a_3 + a_4 + a_5 \]

4. **Calculate the Sum of Coefficients**:
Now, we calculate the sum of the coefficients:
\[ 3 + (-5) + 7 + 4 + (-9) = 3 - 5 + 7 + 4 - 9 = 0 \]
Thus, \( f(1) = 0 \).

5. **Conclusion About Roots**:
Since \( f(1) = 0 \), we conclude that \( x = 1 \) is a root of the polynomial.

6. **Degree of the Polynomial**:
The polynomial \( f(x) \) is a quartic polynomial (degree 4). By the Fundamental Theorem of Algebra, a polynomial of degree \( n \) has exactly \( n \) roots (counting multiplicities).

7. **Determine the Nature of Roots**:
Since \( f(x) \) is a quartic polynomial and we have found one real root (at \( x = 1 \)), it must have at least one more real root (as quartic polynomials can have either 2 or 4 real roots).

8. **Final Statement**:
Therefore, we conclude that the polynomial has at least two real roots.

### Final Answer:
The polynomial \( a_1 x^4 + a_2 x^3 + a_3 x^2 + a_4 x + a_5 = 0 \) has at least two real roots.