The condition that the roots of x^3-b x^2+c x | vedclass

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

Question

(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$. Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3

Vedclass · Accepted Answer

$9 c b=2 b^3+27 d$

Vedclass · Answer

(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$. Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3\alpha$. From the given equation $x^3-b x^2+c x-d=0$,the sum of the roots is $b$. Therefore,$3\alpha = b \Rightarrow \alpha = \frac{b}{3}$. Since $\alpha$ is a root of the equation,it must satisfy $x^3-b x^2+c x-d=0$. Substituting $x = \frac{b}{3}$: $(\frac{b}{3})^3 - b(\frac{b}{3})^2 + c(\frac{b}{3}) - d = 0$ $\frac{b^3}{27} - \frac{b^3}{9} + \frac{bc}{3} - d = 0$ Multiplying the entire equation by $27$: $b^3 - 3b^3 + 9bc - 27d = 0$ $-2b^3 + 9bc - 27d = 0$ $9bc = 2b^3 + 27d$.

The condition that the roots of $x^3-b x^2+c x-d=0$ are in arithmetic progression is

Similar Questions

If the sum of three numbers in an arithmetic sequence is $15$ and the sum of their squares is $83$,then the numbers are

Consider an $A.P.$ of positive integers,whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$. Then its $11^{\text{th}}$ term is:

The number of terms of an $A.P.$ is even; the sum of all the odd terms is $24$,the sum of all the even terms is $30$ and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the $A.P.$ is:

The sum of $3+5+7+\ldots$ to $n$ terms is:

The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is $170$. If there are at least $6$ houses in that row and $a$ is the number of the sixth house,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module