Question

Solve $x^4-5x^3+5x^2+5x-6=0$, given that the product of two of its roots is $3$.

Answer 1

Given equation is $x^4-5x^3+5x^2+5x-6=0$

Let $\alpha ,\beta ,\gamma ,\delta$ be the rootd of the equation.

Then given $\alpha \times \beta =3$

From the equation, $\alpha +\beta +\gamma +\delta =5$ .........(1)

$\alpha \beta +\alpha \delta +\alpha \gamma +\beta \gamma +\beta \delta +\gamma \delta =5$ ..........(2)

$\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =-5$ .........(3)

$\alpha \beta \gamma \delta =-6$ ...........(4)

From $\left(4\right),\gamma \delta =\frac{-6}{3}=-2$

substituting in (2) $\Rightarrow 3+\beta \gamma +\alpha \gamma +\alpha \delta +\beta \delta -2=5$

$(\alpha +\beta )(\gamma +\delta )=4$

Let $\alpha +\beta =m\Rightarrow m(5-m)=4\Rightarrow m=1$ or $4$

Substituting in (3) $\Rightarrow 3\gamma +3\delta +\alpha (-2)+\beta (-2)=-5$

$\Rightarrow 3(\gamma +\delta )-2(\alpha +\beta )=-5$

If $\alpha +\beta =1,\gamma +\delta =-1$ which does not satisfy (1)

$\alpha +\beta =4,\gamma +\delta =1$

and also $\alpha \beta =3,\gamma \delta =-2$

solving these we get the roots as $1,3,2,-1.$