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Question

If for the cubic equation x3−4x2+x+6=0, the possible no. of positive real roots are 2 or 0 and the possible no. of negative real roots is 1, then which of the following can be true?

A
Number of positive roots:0
Number of imaginary roots:3
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B
Number of positive roots:0
Number of imaginary roots:2
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C
Number of positive roots:2
Number of imaginary roots:0
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D
Number of positive roots:2
Number of imaginary roots:3
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Solution

The correct option is C Number of positive roots:2
Number of imaginary roots:0
Given that for the cubic equation x34x2+x+6=0, the possible no. of positive real roots are 2 or 0 and the possible no. of negative real roots is 1.
The total no. of solutions will be 3 as this is a cubic equation.
Also if there are any imaginary roots, then they will come only in conjugate pairs.

Therefore the different possibilities can be shown as:


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