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Q.
Two roots of the cubic $x^3+3 x^2+k x-12=0$ are real and unequal but have the same absolute value. The value of $k$ is
Complex Numbers and Quadratic Equations
Solution:
Let the roots are $r,-r$ and $t$
Hence, $r - r + t =-3 \Rightarrow t =-3$.
Now, product of the roots is $(r)(-r)(t)=12$
$-r^2(-3)=12 \Rightarrow r^2=4$
Hence, roots are $2,-2$ and -3
Now, sum of the product taken two at a time is $r(-r)+(-r) t+t r=k$
$-4-6+6=k \Rightarrow k=-4$.