If the roots alpha, beta, gamma of the equati | vedclass

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

Question

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+px+10=0$ and are in arithmetic progression $(AP)$. Let the roots be $\b

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+px+10=0$ and are in arithmetic progression $(AP)$. Let the roots be $\beta-d, \beta, \beta+d$. From the sum of roots,$(\beta-d) + \beta + (\beta+d) = 6$,which gives $3\beta = 6$,so $\beta = 2$. Since $\beta = 2$ is a root,it must satisfy the equation: $2^3 - 6(2^2) + p(2) + 10 = 0$. $8 - 24 + 2p + 10 = 0$ $\Rightarrow 2p - 6 = 0$ $\Rightarrow p = 3$. The product of the roots is $\alpha\beta\gamma = -10$. Since $\beta = 2$,we have $\alpha\gamma = -5$. Also,$\alpha+\gamma = 6 - 2 = 4$. We use the identity $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))$. Alternatively,$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha))$. Here,$\alpha+\beta+\gamma = 6$,$\alpha\beta+\beta\gamma+\gamma\alpha = p = 3$,and $\alpha\beta\gamma = -10$. $\alpha^3+\beta^3+\gamma^3 - 3(-10) = 6(6^2 - 3(3))$. $\alpha^3+\beta^3+\gamma^3 + 30 = 6(36 - 9) = 6(27) = 162$. $\alpha^3+\beta^3+\gamma^3 = 162 - 30 = 132$.

If the roots $\alpha, \beta, \gamma$ of the equation $x^3-6x^2+px+10=0$ are in arithmetic progression,then $\alpha^3+\beta^3+\gamma^3=$

Similar Questions

Let $x, y, z$ be positive real numbers such that $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$. Then $x^3 + y^3 + z^3$ is equal to

Let the positive integers be written in the form:
$1$
$2$ $3$
$4$ $5$ $6$
$7$ $8$ $9$ $10$
If the $k^{\text{th}}$ row contains exactly $k$ numbers for every natural number $k$,then the row in which the number $5310$ will be,is:

Find the $17^{\text{th}}$ and $24^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = 4n - 3$.

If $a$,$b$,and $c$ are the first,second,and last terms of an arithmetic progression respectively,then the total number of terms is......

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module