Question

Solve 6x3 - 11x2 – 3x + 2 = 0, given that its
roots are in harmonic progression.​

Answer 1

Answer:

Consider the given equation,

6x3−11x2+6x−1=0

Put,x=1 we get

6.13−11.12+6.1−1=0

0=0

Hence, x=1 ⇒x−1=0 is zeroes os given equation.

Now

x−1)6x3−11x2+6x−16x2−5x+1

−(6x3−6x2)

−5x2+6x−1

−(−5x2+5x)

x−1

−(x−1)

Answer 2

The roots of the equation $6x^{3} -11x^{2} -3x+2=0$ are $1,\frac{1}{2}\ and \ \frac {1}{3} .$

Step-by-step explanation:

Given:

The equation $6x^{3} -11x^{2} -3x+2=0$.

The roots are in harmonic progression.

To Find:

The roots of the equation $6x^{3} -11x^{2} -3x+2=0$.

Formula Used:

If

Solution:

Equation $6x^{3} -11x^{2} -3x+2=0$ roots are in harmonic progression.

If we put $x=\frac{1}{y}$  in  the equation $6x^{3} -11x^{2} -3x+2=0$  then roots  will be  in  Arithmetic progression.

Equation will be $6(\frac{1}{y} )^{3} -11(\frac{1}{y} )^{2} -3(\frac{1}{y} )+2=0$

                             $y^{3} -6y^{2} +11y-6=0$ and roots   in A.P.

  Let  roots of equation are a-d.a and a+d which are in A.P.

  $a-d+a+a+d=-\frac{-6}{1}$

  $a-d+a+a+d=6$

  $3a=6$

  $a=2$

$(a-d)a(a+d)=-\frac{-6}{1}$

$(a-d)a(a+d)=6$

Putting the value of a-=2

$(2-d)\times 2\times(2+d)=6$

$(2-d)\times(2+d)=\frac{6}{2}$

$2^{2} -d^{2} =3$

$d^{2} =4-3=1$

$d=1$

The value of a-d,a and a+d will be $(2-1),(2) \ and (\ 2+1)$ or 1,2 and 3.

It mean the value of y=1,2 and 3.

Hence value of x  will be $1,\frac{1}{2}\ and \ \frac {1}{3}$.

Thus,the roots of the equation $6x^{3} -11x^{2} -3x+2=0$ are $1,\frac{1}{2}\ and \ \frac {1}{3} .$

PROJECT CODE#SPJ3