Multiple choice

Let $(a_1, a_2, a_3, a_4, a_5)$ denote a rearrangement of $(3, -5, 7, 4, -9)$, then the equation $a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0$ has

  1. At least two real roots.

  2. All four real roots.

  3. Only imaginary roots.

  4. Two real and two imaginary roots.
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A Correct answer
Explanation

By Descartes' Rule of Signs, the number of sign changes in the coefficients determines the number of positive/negative roots. Given the set of coefficients, there are sign changes, ensuring at least two real roots.

AI explanation

A polynomial of degree 4 with real coefficients can have at most 4 real roots, and any non-real roots must occur in complex conjugate pairs. Thus, the possible number of non-real roots is 0, 2, or 4, meaning the polynomial must have 4, 2, or 0 real roots. Because the leading coefficient a1 and the constant term a5 have opposite signs in all rearrangements of the numbers (3, -5, 7, 4, -9), the product of the roots is always negative. Since a negative product requires an odd number of positive real roots, the equation cannot have zero real roots, proving it must have at least two real roots.

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