Question

Solve the equation $6x^3-11x^2+6x-1=0$ and find the smallest of roots.Given the roots are in harmonical progression.

• A. 1
• B. 1/3
• C. 1/2
• D. 1/4

Answer 1

The correct option is B 1/3

Replacing $x$ by $1/x,$ we get
$6/x^3-11/x^2+6/x-1=0$
or $x^3-6x^2+11x-6=0$
Its roots are in A.P. say $\alpha ,\beta ,\gamma .$
$\therefore 2\beta =\alpha +\gamma$ or $3\beta =\alpha +\beta +\gamma =6$
or $\beta =2$
Hence $(x-2)$ is a factor of above equation.
$\therefore x^3-6x^2+11x-6=(x-2)(x^2-4x+3)=(x-2)(x-3)(x-1).$
$\therefore$ Roots 1,2,3 are in A.P.or $1,\frac{1}{2},\frac{1}{3}$ are in H.P.
Thus smallest root is $\frac{1}{3}$.