Q129 / 200 · AP EAMCET 2001 · Mathematics

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

  • A
    $1$
  • Correct answer
    B
    $2$
  • C
    $\frac{1}{2}$
  • D
    $\frac{1}{3}$

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 $x^2=0$ $\therefore \quad 3 h-6=0 \Rightarrow h=2$

Hence, the correct answer is option: 'B'.

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