Question

The set of values of c for which ${{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}}$ is of the form ${\left( {{\text{x}} - {{\alpha }}} \right)^2}\left( {{\text{x}} - {{\beta }}} \right),\;\left[ {{{\alpha }},\;{{\beta }} \in {\text{R}}} \right]$ is given by-
A. {0}
B. {4}
C. {0, 4}
D. Null set

Answer 1

Hint: A cubic equation is the one whose degree is 3, that is the highest power of x is 3. They have at most three roots and can have a lesser number of roots as well. Here, we have been given the form of the given expression, so we just need to equate these two equations and compare them term by term to get the value(s) of c. Some formulas to be used are-
${\left( {{\text{a}} \pm {\text{b}}} \right)^2} = {{\text{a}}^2} \pm 2ab + {{\text{b}}^2}$...(1)

Complete step-by-step solution -
We have to obtain the value of c from ${{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}}$ such that it has to be of the form ${\left( {{\text{x}} - {{\alpha }}} \right)^2}\left( {{\text{x}} - {{\beta }}} \right),\;\left[ {{{\alpha }},\;{{\beta }} \in {\text{R}}} \right]$. So we will equate the two expressions as-
 ${{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}} = {\left( {{\text{x}} - {{\alpha }}} \right)^2}\left( {{\text{x}} - {{\beta }}} \right)$
Using property (1) we can expand as-
$\begin{align}
  &{{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}} = \left( {{{\text{x}}^2} - 2\alpha x + {{{\alpha }}^2}} \right)\left( {{\text{x}} - {{\beta }}} \right) \\
  &{{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}} = {{\text{x}}^3} - \beta {x^2} - 2\alpha {x^2} + 2\alpha \beta x + {{{\alpha }}^2}{\text{x}} - {\alpha ^2}\beta \\
  &{{\text{x}}^3} - 6{{\text{x}}^2} + 9{\text{x}} - {\text{c}} = {{\text{x}}^3} - \left( {{{\beta }} + 2{{\alpha }}} \right){{\text{x}}^2} + \left( {{{{\alpha }}^2} + 2\alpha \beta } \right){\text{x}} - {\alpha ^2}\beta \\
\end{align} $
We will now compare the coefficients for each of the degrees of x as-
$\begin{align}
  & - 6 = - \left( {{{\beta }} + 2{{\alpha }}} \right) \\
  &2{{\alpha }} + {{\beta }} = 6...\left( 2 \right) \\
  &9 = {{{\alpha }}^2} + 2\alpha \beta \\
  &{{{\alpha }}^2} + 2\alpha \beta = 9...\left( 3 \right) \\
   &- {\text{c}} = - {\alpha ^2}\beta ...\left( 4 \right) \\
\end{align} $
We will first solve equations (2) and (3) as-
$\begin{align}
  &{{\beta }} = 6 - 2{{\alpha }} \\
 &{{{\alpha }}^2} + 2\alpha \beta = 9 \\
 & Subsituting\;the\;value\;of\;{{\beta }}, \\
  &{{{\alpha }}^2} + 2{{\alpha }}\left( {6 - 2{{\alpha }}} \right) = 9 \\
  &{{{\alpha }}^2} + 12{{\alpha }} - 4{{{\alpha }}^2} = 9 \\
   &- 3{{{\alpha }}^2} + 12{{\alpha }} - 9 = 0 \\
  &{{{\alpha }}^2} - 4{{\alpha }} + 3 = 0 \\
  &{{{\alpha }}^2} - 3{{\alpha }} - {{\alpha }} + 3 = 0 \\
  &{{\alpha }}\left( {{{\alpha }} - 3} \right) - 1\left( {{{\alpha }} - 3} \right) = 0 \\
  &\left( {{{\alpha }} - 3} \right)\left( {{{\alpha }} - 1} \right) = 0 \\
  &{{\alpha }} = 1,\;3 \\
\end{align} $
Subsequently, we get the values of as-
$\begin{align}
  &{{\beta }} = 6 - 2{{\alpha }} \\
  &{{\beta }} = 6 - 2\left( 3 \right),\;6 - 2\left( 1 \right) \\
  &{{\beta }} = 0,4 \\
\end{align} $
Using equation (4),
$\begin{align}
  &{\text{c}} = {{{\alpha }}^2}{{\beta }} \\
  &{\text{c}} = {3^2}\left( 0 \right)\;or\;{\text{c}} = {\left( 1 \right)^2} \times 4 \\
  &{\text{c}} = 0\;or\;4 \\
\end{align} $
Hence the set of values of c is {0, 4}. The correct option is C.

Note: In such types of questions involving quadratic, cubic or higher equations, more than one value may be possible, so we should keep in mind that we use all the possible values. Students often use and write only one of the values, which leads to an incorrect answer. Also, we should remember the basic algebraic formulas to solve these problems.