Question

Let $f\left(x\right)=x^3+p{x}^2+qx+12$ where $p,q$ are integers. If $f\left(x\right)=0$ has a pair of real roots of equal magnitude and opposite sign, then the number of possible pairs of values $(p,q)$ is

• A. $0$
• B. $1$
• C. $6$
• D. $12$

Answer 1

The correct option is C $6$

$f\left(x\right)=x^3+p{x}^2+qx+12$
Let $\alpha ,-\alpha ,\beta$ be the roots of the equation.
Then $\alpha +(-\alpha )+\beta =-p$
$-{\alpha }^2+(-\alpha \beta )+\beta \alpha =q$
$-{\alpha }^2\beta =-12$

$\Rightarrow \beta =-p\cdots \left(1\right)$
$-{\alpha }^2=q\cdots \left(2\right)$
$-{\alpha }^2\beta =-12\cdots \left(3\right)$

From $\left(1\right),\left(2\right)\text{and}\left(3\right),$ we get
$pq=12$
We can verify $p,q$ both cannot be positive by Descartes' rule.
$\therefore$ The possible values of $(p,q)$ are
$(-1,-12),(-2,-6),(-3,-4),$
$(-12,-1),(-6,-2),(-4,-3)$