Consider equation for $\lambda\neq 0$, $$x^5+5\lambda x^4-x^3+(\alpha\lambda-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0.$$ If $p, q$ are values of $\alpha$ for which equation has exactly one root $\in$ $\mathbf C$ independent of $\lambda$ and exactly two roots $\in$ $\mathbf C$ independent of $\lambda$ then find $p, q$.

Where $\mathbf C$ denotes set of complex numbers.

I don't understand how can a polynomial have exactly one complex root, I think that complex roots of any odd degree polynomial are always found in conjugate pairs.

  • 1
    $\begingroup$ Complex roots must come in conjugate pairs when the coefficients are all real. The only condition on $\lambda$ that you’ve given is that it’s nonzero. $\endgroup$
    – amd
    Feb 4 '18 at 19:20
  • $\begingroup$ Note, a real root independent of $\lambda$ also is $\in \mathbb C$. $\endgroup$
    – Macavity
    Feb 5 '18 at 1:48

Hint:  write the equation as:

$$(x^5-x^3-4x^2-3x-2) + \lambda(5 x^4+\alpha x^2-8 x+\alpha)=0$$

Any root which is independent of $\,\lambda\,$ must be a common root of:

$$ \begin{cases} \begin{align} x^5-x^3-4x^2-3x-2 &= 0\\ 5 x^4+\alpha x^2-8 x+\alpha &= 0 \end{align} \end{cases} $$

Note that the first of those factors nicely, and has one simple root and two double ones.


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