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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation $x^3 - 10x^2 + 7x + 8 = 0$.From the relation between roots and coefficients:$\alpha + \beta + \gamma = 10$ ($A$-$5$)$\alpha\bet

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the equation $x^3 - 10x^2 + 7x + 8 = 0$.From the relation between roots and coefficients:$\alpha + \beta + \gamma = 10$ ($A$-$5$)$\alpha\beta + \beta\gamma + \gamma\alpha = 7$$\alpha\beta\gamma = -8$Now,$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (10)^2 - 2(7) = 100 - 14 = 86$ ($B$-$3$)Next,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{7}{-8} = -\frac{7}{8}$ ($C$-$2$)Finally,$\frac{\alpha}{\beta\gamma} + \frac{\beta}{\gamma\alpha} + \frac{\gamma}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta\gamma} = \frac{86}{-8} = -\frac{43}{4}$ ($D$-$1$)Thus,the correct matching is $A-5, B-3, C-2, D-1$.

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

Similar Questions

If $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$,then the value of $\frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$ is

If the ratio of the roots of the equation $12x^2 - mx + 5 = 0$ is $2 : 3$,then $m = .....$

The roots of the equation $x^2 + bx - c = 0$ where $b, c > 0$ are:

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is:

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$,then the equation whose roots are $\alpha+\beta$ and $\frac{1}{\alpha}+\frac{1}{\beta}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module