Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} ε R$ denote a rearrangement of equation $p_{1} x^{5} + p_{2} x^{3} + p_{3} x^{2} + p_{4} x + p_{5} = 0$ then, 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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none of these

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Solution

The correct option is A at least two real roots
$a_{1} x^{4} + a_{2} x^{3} + a_{3} x^{2} + a_{4} x + a_{5} = 0$
for $x = 1$
$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 0$
for the given set of equation,
sum of $p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 0$
So, $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} ϵ R$ for any set of arrangement
Hence at least two real roots.

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Let a1,a2,a3,a4,a5εR denote a rearrangement of equation p1x5+p2x3+p3x2+p4x+p5=0 then, equation a1x4+a2x3+a3x2+a4x+a5=0 has

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