Find the least number of imaginary roots in $x^{5} - 7 x^{3} + 3 x^{2} + 1 = 0$

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Solution

The correct option is D 2
$x^{5} - 7 x^{3} + 3 x^{2} + 1 = 0$
$f (x) = x^{5} - 7 x^{3} + 3 x^{2} + 1$
+ - + +
Maximum number of positive roots = 2
$f (- x) = - x^{5} + 7 x^{3} + 3 x^{2} + 1$
- + + +
Maximum number of negative roots = 1
Maximum number of real roots =3
Total roots of the equation =5
So, least number of complex roots 5-3 =2

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