alpha, beta, gamma are the roots of the equat | vedclass

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Question

(A) Given the cubic equation $x^3-10x^2+7x+8=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas:$\alpha+\beta+\gamma = -(-10)/1 = 10$ ($A$-

Vedclass · Accepted Answer

$A-5, B-3, C-2, D-1$

Vedclass · Answer

(A) Given the cubic equation $x^3-10x^2+7x+8=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas:$\alpha+\beta+\gamma = -(-10)/1 = 10$ ($A$-$5$)$\alpha\beta+\beta\gamma+\gamma\alpha = 7/1 = 7$$\alpha\beta\gamma = -8/1 = -8$For $B$: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (10)^2 - 2(7) = 100 - 14 = 86$ ($B$-$3$)For $C$: $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma} = \frac{7}{-8} = -\frac{7}{8}$ ($C$-$2$)For $D$: $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta} = \frac{\alpha^2+\beta^2+\gamma^2}{\alpha\beta\gamma} = \frac{86}{-8} = -\frac{43}{4}$ ($D$-$1$)Thus,the correct matching is $A-5, B-3, C-2, D-1$.

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

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