Question

If the product of two roots of the equation $x^4-5x^3+5x^2+5x-6=0$ is $3$, then which of the following is/are correct?

• A. The equation has only one negative root.
• B. The equation has three negative roots.
• C. The product of all positive roots will be $6.$
• D. The product of all negative roots will be $6.$

Answer 1

Answer: (A), (C)

The correct options are
A The equation has only one negative root.
C The product of all positive roots will be $6.$

Let $\alpha ,\beta ,\gamma ,\delta$ be the roots of the equation given, so
$s_1=\alpha +\beta +\gamma +\delta =5$
$s_2=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =5$
$s_3=\alpha \beta \gamma +\alpha \beta \delta +\gamma \delta \alpha +\gamma \delta \beta =-5$
$s_4=\alpha \beta \gamma \delta =-6$

Assuming $\alpha \beta =3$ then from $s_4$
$\gamma \delta =-2$
Now from $s_3$, we get
$3(\gamma +\delta )-2(\alpha +\beta )=-5$
From $s_1$
$(\gamma +\delta )+(\alpha +\beta )=5$
Solving these two equations on $\alpha +\beta ,\gamma +\delta$, we get
$\alpha +\beta =4,\gamma +\delta =1$

Now,
$\alpha +\beta =4,\alpha \beta =3$ and $\gamma +\delta =1,\gamma \delta =-2$

Solving them, we get
$\alpha =3,\beta =1\text{or}\alpha =1,\beta =3$
$\gamma =2,\delta =-1\text{or}\gamma =-1,\delta =2$

Therefore, the roots of the given equation are $-1,1,2,3$.