If the roots of the equation 6x^3-11x^2+6x-1= | vedclass

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

Question

(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmeti

Vedclass · Accepted Answer

Arithmetic Progression

Vedclass · Answer

(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmetic progression.Substituting $x = \frac{1}{y}$ in the given equation,we get $6(\frac{1}{y})^3 - 11(\frac{1}{y})^2 + 6(\frac{1}{y}) - 1 = 0$,which simplifies to $y^3 - 6y^2 + 11y - 6 = 0$.Now,consider the equation $x^3-6x^2+11x-6=0$.Since $x=1$ satisfies the equation $(1-6+11-6=0)$,$(x-1)$ is a factor.Dividing by $(x-1)$,we get $(x-1)(x^2-5x+6)=0$,which factors to $(x-1)(x-2)(x-3)=0$.The roots are $1, 2, 3$.Since $2-1 = 1$ and $3-2 = 1$,the roots are in Arithmetic Progression.

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in

Similar Questions

If the sum of three numbers in an arithmetic progression is $33$ and their product is $792$,what is the smallest number among them?

Let $A, G, H$ and $S$ respectively denote the arithmetic mean,geometric mean,harmonic mean and the sum of the numbers $a_1, a_2, a_3, \ldots, a_n$. Then the value of $x$ at which the function $f(x)=\sum_{k=1}^n(x-a_k)^2$ has a minimum is

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,the number of terms is

For $n \in N$,what is the value of $1+4+7+\cdots+(3n-2)$?

The sum of all two-digit positive numbers which,when divided by $7$,yield $2$ or $5$ as a remainder is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module