Question

The roots of the equation $x^4+x^3+2x-4=0$ are $\alpha ,\beta ,\gamma$ and $\delta$, such that $|\alpha |\le |\beta |\le |\gamma |\le |\delta |$. If the sum of two roots is zero, then which of the following is/are correct?

• A. Two real roots and two imaginary roots.
• B. $\alpha ,\beta ,\gamma \text{are in G.P.}$
• C. $\alpha ,\beta ,\delta \text{are in G.P.}$
  
• D. $\alpha ,\gamma ,\delta \text{are in G.P.}$

Answer 1

Answer: (A), (C), (D)

$\alpha ,\gamma ,\delta \text{are in G.P.}$

$x^4+x^3+2x-4=0$
$\alpha +\beta +\gamma +\delta =-1$
$\text{Let}\alpha +\beta =0$
$\therefore \gamma +\delta =-1$
Quadratic factor corresponding to $\alpha ,\beta$
$x^2-0.x+p=x^2+p$
Corresponding to $\gamma ,\delta$
$x^2+x+q$

$\therefore x^4+x^3+2x-4=(x^2+p)(x^2+x+q)\Rightarrow x^4+x^3+2x-4=x^4+x^3+x^2(p+q)+px+pq$
Comparing the coefficients of $x^2$ and $x$, we get
$p=2,q=-2$
So,
$\Rightarrow (x^2+2)(x^2+x-2)=0\Rightarrow (x-1)(x+2)(x^2+2)=0\Rightarrow x=1,-2,\pm i\sqrt{2}$

$\alpha =1,\beta =-i\sqrt{2},\gamma =i\sqrt{2},\delta =-2\because |\alpha |\le |\beta |\le |\gamma |\le |\delta |$
$\Rightarrow \alpha ,\beta ,\delta \text{are in G.P.}$
$\Rightarrow \alpha ,\gamma ,\delta \text{are in G.P.}$