Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3-10x^2+7x+8=0$

Match the elements of list I with elements of list II:

List I	List II
P) $\alpha +\beta +\gamma$	a) $\frac{43}{4}$
Q) ${\alpha }^2+{\beta }^2+{\gamma }^2$	b) $-\frac{7}{8}$
R) $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$	c) $86$
S) $\frac{\alpha }{\beta \gamma }+\frac{\beta }{\gamma \alpha }+\frac{\gamma }{\alpha \beta }$	d) $0$
	e) $10$

Then correct arrangement of P, Q, R, S is respectively

• A. $e,c,a,b$
• B. $d,c,a,b$
• C. $e,c,b,a$
• D. $e,b,c,a$

Answer 1

The correct option is C $e,c,b,a$

Given: $\alpha ,\beta$ and $\gamma$ are the roots of $x^3-10x^2+7x+8=0$
Using relation between roots and coefficient, we have
$\alpha +\beta +\gamma =10\alpha \beta +\beta \gamma +\gamma \alpha =7\alpha \beta \gamma =-8$
Now,
$\alpha +\beta +\gamma =10$ which is $e$
${\alpha }^2+{\beta }^2+{\gamma }^2={(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )=100-2\left(7\right)=86$ which is $c$
$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }=\frac{7}{-8}=\frac{-7}{8}$ which is $b$
$\frac{\alpha }{\beta \gamma }+\frac{\beta }{\alpha \gamma }+\frac{\gamma }{\alpha \beta }=\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{\alpha \beta \gamma }$

$=\frac{{(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )}{\alpha \beta \gamma }=\frac{100-14}{-8}=\frac{43}{4}$ which is $a$.
Hence, the answer is $e,c,b,a$.