Each of the roots of the equation $x^3-6 x^2+6 x-5=0$ are increased by $h$. So that the new transformed…
Each of the roots of the equation $x^3-6 x^2+6 x-5=0$ are increased by $h$. So that the new transformed equation does not contain $x^2$ term, then $h$ is equal to
$1$
$2$
$\frac{1}{2}$
$\frac{1}{3}$
Solution
Given equation is
$x^3-6 x^2+6 x-5=0$
$x$ of this equation replaced by $x+h$.
$\begin{gathered}\therefore(x+h)^3-6(x+h)^2+6(x+h)-5=0 \\ x^3+h^3+3 x^2 h+3 x h^2-6\left(x^2+h^2+2 x h\right) \\ +6 x+6 h-5=0\end{gathered}$
Since, coefficient of $x^2=0$
$\therefore \quad 3 h-6=0 \Rightarrow h=2$