Question

If the product of two of the roots $x^4-8x^3+21x^2-20x+5=0$ is $5$, then the roots are

• A. $\frac{2\pm \sqrt{5}}{2},\frac{5\pm \sqrt{5}}{2}$
• B. $\frac{3\pm \sqrt{5}}{2},\frac{5\pm \sqrt{5}}{2}$
• C. $5,1\pm \sqrt{2}$
• D. $\frac{4\pm \sqrt{5}}{2},\frac{5\pm \sqrt{5}}{2}$

Answer 1

Answer: (B)

$\frac{3\pm \sqrt{5}}{2},\frac{5\pm \sqrt{5}}{2}$

Let $\alpha ,\beta ,\gamma ,\delta$ be the roots of the equation $x^4-8x^3+21x^2-20x+5=0$,

where $\alpha \times \beta =5$

Now, $S_1=\alpha +\beta +\gamma +\delta =8$ ...(1)

$S_2=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =21$

$\Rightarrow \alpha \beta +(\alpha +\beta )(\gamma +\delta )+\gamma \delta =21$ ...(2)

$S_3=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =20$

$\Rightarrow \alpha \beta (\gamma +\delta )+\gamma \delta (\alpha +\beta )=20$

$\Rightarrow 5(\gamma +\delta )+\gamma \delta (\alpha +\beta )=20$ ...(3)

$S_4=\alpha \beta \gamma \delta =5$

$\Rightarrow 5\gamma \delta =5\Rightarrow \gamma \delta =1$ ...(4)

Substituting this in (3), we get

$5(\gamma +\delta )+(\alpha +\beta )=20$ ...(5)

Now solving (1) and (5), we get

$4(\gamma +\delta )=12\Rightarrow \gamma +\delta =3$

As ${(\gamma -\delta )}^2={(\gamma +\delta )}^2-4\gamma \delta$

$=9-4=5$

$\Rightarrow \gamma -\delta =\pm \sqrt{5}$

Hence, from the given options we can conclude that $\gamma ,\delta =\frac{3\pm \sqrt{5}}{2}$ and $\alpha ,\beta =\frac{5\pm \sqrt{5}}{2}$