Question
The number of values of $$k$$ for which $$\left\{ {{x^2} - \left( {k - 2} \right)x + {k^2}} \right\}\left\{ {{x^2} + kx + \left( {2k - 1} \right)} \right\}$$ is a perfect square is
A.
1
B.
2
C.
0
D.
None of these
Answer :
1
Solution :
$${x^2} - \left( {k - 2} \right)x + {k^2} = 0\,\,{\text{and }}{x^2} + kx + 2k - 1 = 0$$ should have both roots common or each should have equal roots.
$$\eqalign{
& \therefore \,\,\left( {\text{i}} \right)\frac{1}{1} = \frac{{ - \left( {k - 2} \right)}}{k} = \frac{{{k^2}}}{{2k - 1}} \cr
& {\text{or, }}\left( {{\text{ii}}} \right){\left( {k - 2} \right)^2} - 4{k^2} = 0\,\,{\text{and }}{k^2} - 4\left( {2k - 1} \right) = 0. \cr
& \left( {\text{i}} \right) \Rightarrow \,\,k = - k + 2,2k - 1 = {k^2}.\,{\text{So, }}k = 1 \cr
& \left( {{\text{ii}}} \right)\, \Rightarrow \,\,\left( {3k - 2} \right)\left( { - k - 2} \right) = 0,{k^2} - 8k + 4 = 0.\,\,{\text{So, no value of }}k{\text{ is possible}}{\text{.}} \cr} $$