Let $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ denote a rearrangement of $(3, - 5, 7, 4, - 9)$ , then the equation $a_{1} x^{4} + a_{2} x^{3} + a_{3} x^{2} + a_{4} x + a_{5} = 0$ has

At least two real roots.

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All four real roots.

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Only imaginary roots.

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Two real and two imaginary roots.

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Solution

The correct option is A At least two real roots.
$3 + (- 5) + 7 + 4 + (- 9)$
$= 14 - 14$
$= 0$
Hence,
$x = 1$ will always be one of the roots.
Now, there are total four roots and we know that complex roots appear in pairs.
However, $x = 1$ is already a real root.
Hence, out of the remaining $3$ , roots, maximum of $2$ can be imaginary.
Hence, atleast $2$ will be real roots always.

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