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Question

Let f(x)=x3+px2+qx+12 where p,q are integers. If f(x)=0 has a pair of real roots of equal magnitude and opposite sign, then the number of possible pairs of values (p,q) is

A
0
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B
1
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C
6
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D
12
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Solution

The correct option is C 6
f(x)=x3+px2+qx+12
Let α,α,β be the roots of the equation.
Then α+(α)+β=p
α2+(αβ)+βα=q
α2β=12

β=p (1)
α2=q (2)
α2β=12 (3)

From (1),(2) and (3), we get
pq=12
We can verify p,q both cannot be positive by Descartes' rule.
The possible values of (p,q) are
(1,12),(2,6),(3,4),
(12,1),(6,2),(4,3)


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